1) The first column of the transition matrix is (a1, a2, a3) = (0, 0, 1).
2) The second column of the transition matrix is (b1, b2, b3) = (0, 1, 0).
3) The third column of the transition matrix is (c1, c2, c3) = (1, 0, 0).
To find the transition matrix from basis b to basis b', we need to express each vector in b' as a linear combination of vectors in b and then arrange the coefficients in a matrix.
Let's start with the first vector in b', (0, 0, 1):
(0, 0, 1) = a1(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1)
Simplifying this equation, we get:
a1 = 0
a2 = 0
a3 = 1
Therefore, the first column of the transition matrix is (a1, a2, a3) = (0, 0, 1).
Now let's move on to the second vector in b', (0, 1, 0):
(0, 1, 0) = b1(1, 0, 0) + b2(0, 1, 0) + b3(0, 0, 1)
Simplifying this equation, we get:
b1 = 0
b2 = 1
b3 = 0
Therefore, the second column of the transition matrix is (b1, b2, b3) = (0, 1, 0).
Finally, let's look at the third vector in b', (1, 0, 0):
(1, 0, 0) = c1(1, 0, 0) + c2(0, 1, 0) + c3(0, 0, 1)
Simplifying this equation, we get:
c1 = 1
c2 = 0
c3 = 0
Therefore, the third column of the transition matrix is (c1, c2, c3) = (1, 0, 0).
Putting it all together, we get the transition matrix from basis b to basis b':
| 0 0 1 |
| 0 1 0 |
| 1 0 0 |
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Depths of pits on a corroded steel surface are normally distributed with mean 818 μm and standard deviation 29 μm.
A) Find the 10th percentile of pit depths.
B) A certain pit is 780 μm deep. What percentile is it on? (Round up the final answer to the nearest whole number.)
C) What proportion of pits have depths between 800 and 830 μm?
The 10th percentile of pit depths is 780μm. A certain pit with 780 μm deep is at the 10th percentile. The proportion of pits have depths between 800 and 830 μm is 7.33%.
A)
To find the 10th percentile of pit depths, we need to use the z-score table. Where x = μ + zσ, here we are looking for the z-score, for the given 10th percentile.
Using the standard normal distribution table, we get the value of -1.28 which corresponds to the 10th percentile.
Therefore,
x = 818 - 1.28 * 29x = 779.88 = 780μm.
So, 780μm is the 10th percentile of pit depths.
B)
We are given that the mean is 818 μm and standard deviation is 29 μm. A certain pit is 780 μm deep. To find the percentile for this, we need to find the z-score for this given pit.
x = 780 μm, μ = 818 μm, σ = 29 μm
Now, z-score can be found as,
z = (x - μ) / σ = (780 - 818) / 29 = -1.31
We can find the percentile using the standard normal distribution table.
Therefore, the given pit is at the 10th percentile.
C)
We are given that the mean is 818 μm and standard deviation is 29 μm. The proportion of pits with depths between 800 and 830 μm can be calculated as follows:
P(z < (X- x) / σ) - P(z < (830 - 818) / 29) - P(z < (800 - 818) / 29)
P(z < -0.41) - P(z < -0.62) = 0.3409 - 0.2676 = 0.0733
(rounded off to four decimal places)
Therefore, approximately 7.33% of pits have depths between 800 and 830 μm.
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You want to approximate the area between the graph of f(x)= square root of x
and the x-axis from x=2 to x=6 using a Riemann sum with 30 rectangles. The right endpoint Riemann sum in sigma notation is: ∑ A i=1 B, where A= B= Hint: for B your answer should be an expression involving i.
Riemann Sum = ∑ [f(2 + iΔx)] Δx (when i = 0 to 30)
Given function is f(x) = √x
We want to find the area between x = 2 and x = 6 using right endpoint Riemann sum with 30 rectangles.
The width of each rectangle = Δx= (6-2)/30= 0.1333
B = Right endpoints of subintervals =(2 + iΔx), where i = 0, 1, 2, ... , 30
A = Area between f(x) and x-axis for each subinterval.
Ai = [f(2 + iΔx)] Δx
∴ Riemann Sum = ∑ Ai=1 30∑ [f(2 + iΔx)] Δx
∴ Riemann Sum = ∑ [f(2 + iΔx)] Δx (when i = 0 to 30)
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problems 10−14, given the parent function and a description of the transformation, write the equation of th transformed function, f(x). 10. Absolute value-vertical shift down 5 , horizontal shift right 3 ..................................... 11. Linear-vertical shift up 5.......................... . 12. Square Root -vertical shift down 2 , horizontal shift left 7............................... 13. Quadratic-horizontal shift left 8................................... 14. Quadratic-vertex at (−5,−2)..............................
The transformed absolute value function has a vertical shift down 5 and a horizontal shift right 3. Its equation is f(x) = |x - 3| - 5.
The transformed linear function has a vertical shift up 5. Its equation is f(x) = x + 5.
The transformed square root function has a vertical shift down 2 and a horizontal shift left 7. Its equation is f(x) = √(x + 7) - 2.
The transformed quadratic function has a horizontal shift left 8. Its equation is f(x) = (x + 8)^2.
The transformed quadratic function has a vertex at (-5, -2). Its equation is f(x) = (x + 5)^2 - 2.
For the absolute value function, shifting it down 5 units means subtracting 5 from the function, and shifting it right 3 units means subtracting 3 from the input value. Thus, the transformed equation is f(x) = |x - 3| - 5.
For the linear function, shifting it up 5 units means adding 5 to the function. Therefore, the equation of the transformed function is f(x) = x + 5.
For the square root function, shifting it down 2 units means subtracting 2 from the function, and shifting it left 7 units means subtracting 7 from the input value. Hence, the transformed equation is f(x) = √(x + 7) - 2.
For the quadratic function, shifting it left 8 units means subtracting 8 from the input value. Therefore, the equation of the transformed function is f(x) = (x + 8)^2.
For the quadratic function, having a vertex at (-5, -2) means the vertex of the parabola is located at that point. The equation of the transformed function can be obtained by shifting the standard quadratic equation f(x) = x^2 to the left 5 units and down 2 units. Thus, the equation is f(x) = (x + 5)^2 - 2.
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manny swam x laps at the pool on monday. on tuesday he swam 6 laps more than what he swam on monday. how many laps did he swim on tuesday? how many laps did he swim on both days combined?
The total number of laps would be [tex]"x + (x + 6)"[/tex]. We cannot determine the specific number of laps Manny swam on either day or the total number of laps without this information.
To find out how many laps Manny swam on Tuesday, we need to know the number of laps he swam on Monday.
Let's assume he swam "x" laps on Monday.
On Tuesday, Manny swam 6 laps more than what he swam on Monday.
Therefore, the number of laps he swam on Tuesday would be [tex]"x + 6".[/tex]
To find out how many laps Manny swam on both days combined, we simply add the number of laps he swam on Monday and Tuesday.
So the total number of laps would be[tex]"x + (x + 6)".[/tex]
Please note that the exact value of "x" is not provided in the question, so we cannot determine the specific number of laps Manny swam on either day or the total number of laps without this information.
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On Monday, Manny swam x laps at the pool. On Tuesday, he swam 6 laps more than what he swam on Monday. Manny swam 2x + 6 laps on both Monday and Tuesday combined.
To find out how many laps Manny swam on Tuesday, we need to add 6 to the number of laps he swam on Monday.
Therefore, the number of laps Manny swam on Tuesday can be expressed as (x + 6).
To determine how many laps Manny swam on both days combined, we add the number of laps he swam on Monday to the number of laps he swam on Tuesday.
Thus, the total number of laps Manny swam on both days combined is (x + x + 6).
To simplify this expression, we can combine the like terms:
2x + 6
Therefore, Manny swam 2x + 6 laps on both Monday and Tuesday combined.
In summary, Manny swam (x + 6) laps on Tuesday and 2x + 6 laps on both days combined.
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reduce to row echelon form and solve the system. 2x 4y-2z=2 4x 9y-3z=8 -2x-3y 7z=10
To reduce the given system to row echelon form and then solve it, we need to follow these steps:Write the given system in the matrix form, then represent the system in the augmented matrix form.the solution of the given system is [tex]x = 0, y = 2, and z = 2.[/tex]
Apply the elementary row operations to get the matrix in echelon form.Then apply back substitution to solve the system.Let's solve the given system of equations by the above-mentioned method:First, we represent the system in the matrix form as:[tex]$$\begin{bmatrix}2 & 4 & -2\\4 & 9 & -3\\-2 & -3 & 7\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}2\\8\\10\end{bmatrix}$$[/tex]
Then, we represent the augmented matrix as:[tex]$$\begin{bmatrix}[ccc|c]2 & 4 & -2 & 2\\4 & 9 & -3 & 8\\-2 & -3 & 7 & 10\end{bmatrix}$$[/tex]
Therefore, the row echelon form of the given system is[tex]$$\begin{bmatrix}[ccc|c]1 & 0 & 0 & 0\\0 & 1 & 0 & 2\\0 & 0 & 4 & 8\end{bmatrix}$$[/tex]
Now, applying back substitution, we get the value of z as:[tex]$$4z = 8 \Rightarrow z = \frac 82 = 2$$[/tex]
Next, using z = 2 in the second row of the echelon form,
we get the value of y as:[tex]$$y = 2$$[/tex]
Finally, using [tex]z = 2 and y = 2[/tex]in the first row of the echelon form,
we get the value of x as:[tex]$$x = 0$$[/tex]
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The problem describes a sebe to be amsitized. (Round your anawers to the nearest cent.) in 17\%, compounded semiannuaily. (a) Find the she of each pwiment. $________ (b)Fad the tutal amsurt paid for the purchase. $________ (e) Find the totst interest pald over the life of the ban. $________
Given the problem describes a scene to be amortized, 17% compounded semiannually.
We have to find the she of each payment, the total amount paid for the purchase and the total interest paid over the life of the loan.
(a) Find the of each payment amortization schedule for the given problem shown below: Semiannual Payment $1618.63
(b) Find the total amount paid for the purchase.
Using the formula for the present value of an annuity, the total amount paid for the purchase is:$93,348.80
(c) Find the total interest paid over the life of the loan.
To find the total interest paid over the life of the loan, we need to find the difference between the total amount paid and the amount financed.$93,348.80 - $80,000 = $13,348.80
Therefore, the total interest paid over the life of the loan is $13,348.80.
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The size of each payment is $731.67.
The total amount paid for the purchase is $29,266.80
The total interest paid over the life of the loan is $4,266.80.
Given that a problem describes a Sebe to be amortized in 17%, compounded semi-annually.
We need to find
(a) Find the size of each payment.
(b) Find the total amount paid for the purchase.
(c) Find the total interest paid over the life of the loan.
(a) Size of each payment:
The formula to find the size of each payment is given as:
PV = PMT x [1 - (1 / (1 + r) ^ n)] / r
PV = $25,000
PMT = ?
r = 8.5% / 2 = 4.25% (because compounded semi-annually)
n = 20 x 2 = 40 months
Using the above formula, we get:
25000 = PMT x [1 - (1 / (1 + 0.0425) ^ 40)] / 0.0425
PMT = $731.67
So, the size of each payment is $731.67.
(b) Total amount paid for the purchase:
The total amount paid for the purchase is calculated by multiplying the size of each payment by the total number of payments.
The total number of payments is 20 x 2 = 40 months
The total amount paid = $731.67 x 40 = $29,266.80
So, the total amount paid for the purchase is $29,266.80
(c) Total interest paid over the life of the loan:
The formula to find the total interest paid over the life of the loan is given as:
Total interest = Total amount paid - Amount borrowed
Total amount paid = $29,266.80
Amount borrowed = $25,000
Total interest = $29,266.80 - $25,000 = $4,266.80
So, the total interest paid over the life of the loan is $4,266.80.
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Let F be the radial force field F=xi+yj. Find the work done by this force along the following two curves, both which go from (0,0) to (6,36). (Compare your answers!) A. If C 1
is the parabola: x=t,y=t 2
,0≤t≤6, then ∫ C 1
F⋅dr= B. If C 2
is the straight line segment: x=6t 2
,y=36t 2
,0≤t≤1, then ∫ C 2
F⋅dr=
The force field is given by F = xi + yj. The parabola goes from (0, 0) to (6, 36) and is parameterized as r(t) = ti + t^2j, where 0 ≤ t ≤ 6, and r'(t) = i + 2tj.
The work done by the force along the curve C1 is given by the integral below:∫ C 1 F.dr = ∫ 0 6 F(r(t)).r'(t)dt= ∫ 0 6 (ti + t^3j).(i + 2tj) dt= ∫ 0 6 (t + 2t^4) dt= (t^2/2 + 2t^5/5) [0,6]= 252
The straight line segment goes from (0, 0) to (6, 36) and is parameterized as r(t) = 6ti + 36tj, where 0 ≤ t ≤ 1, and r'(t) = 6i + 36j.
The work done by the force along the curve C2 is given by the integral below:∫ C 2 F.dr = ∫ 0 1 F(r(t)).r'(t)dt= ∫ 0 1 (6ti + 36tj).(6i + 36j) dt= ∫ 0 1 (36t + 216t) dt= (126t^2) [0,1]= 126
Therefore, the answer to the problem is: A. If C1 is the parabola: x = t, y = t^2, 0 ≤ t ≤ 6, then ∫C1 F.dr = 252.B.
If C2 is the straight line segment: x = 6t, y = 36t, 0 ≤ t ≤ 1, then ∫C2 F.dr = 126.
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Change the power series so that it contains x^n.
1. x^(n -1) =____
2. x^(n -2) =____
To change the power series to contain x^n, we can manipulate the given terms as follows: 1. x^(n-1) = x^n / x, and 2. x^(n-2) = x^n / (x^2).
To rewrite the power series in terms of x^n, we can manipulate the given terms by using properties of exponents.
1. x^(n-1):
We start with the given term x^(n-1) and rewrite it as x^n multiplied by x^(-1). Using the rule of exponentiation, x^(-1) is equal to 1/x. Therefore, x^(n-1) can be expressed as x^n multiplied by 1/x, which simplifies to x^n / x.
2. x^(n-2):
Similarly, we begin with the given term x^(n-2) and rewrite it as x^n multiplied by x^(-2). Applying the rule of exponentiation, x^(-2) is equal to 1/(x^2). Hence, x^(n-2) can be represented as x^n multiplied by 1/(x^2), which further simplifies to x^n / (x^2).
By manipulating the given terms using exponent properties, we have successfully expressed x^(n-1) as x^n / x and x^(n-2) as x^n / (x^2), thus incorporating x^n into the power series.
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Solve the logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expression. 9 ln(2x) = 36 Rewrite the given equation without logarithms. Do not solve for x. Solve the equation. What is the exact solution? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is (Type an exact answer in simplified form. Use integers or fractions for any numbers in the expression.) B. There are infinitely many solutions. C. There is no solution. What is the decimal approximation to the solution? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is (Type an integer or decimal rounded to two decimal places as needed.) B. There are infinitely many solutions. C. There is no solution.
Given equation is: 9 \ln(2x) = 36, Domain: (0, ∞). We have to rewrite the given equation without logarithms.
Do not solve for x. Let's take a look at the steps to solve the logarithmic equation:
Step 1:First, divide both sides of the equation by 9. \frac{9 \ln(2x)}{9}=\frac{36}{9} \ln(2x)=4
Step 2: Rewrite the equation in exponential form. e^{(\ln(2x))}=e^4 2x=e^4.
Step 3: Solve for \frac{2x}{2}=\frac{e^4}{2}x=\frac{e^4}{2}x=\frac{54.598}{2}x=27.299. We have found the exact solution. So the correct option is:A.
The solution set is \left\{27.299\right\}The given equation is: 9 \ln(2x) = 36. The domain of the logarithmic function is (0, ∞). First, we divide both sides of the equation by 9. This gives us:\frac{9 \ln(2x)}{9}=\frac{36}{9}\ln(2x)=4Now, let's write the equation in exponential form. We have: e^{(\ln(2x))}=e^4. Now solve for x. We get:2x=e^4\frac{2x}{2}=\frac{e^4}{2}x=\frac{e^4}{2}x=\frac{54.598}{2}x=27.299. We have found the exact solution. So the correct option is:A.
The solution set is \left\{27.299\right\}The decimal approximation of the solution is 27.30 (rounded to two decimal places).Therefore, the solution set is \left\{27.299\right\}and the decimal approximation is 27.30. Given equation is 9 \ln(2x) = 36. The domain of the logarithmic function is (0, ∞). After rewriting the equation in exponential form, we get x=\frac{e^4}{2}. The exact solution is \left\{27.299\right\} and the decimal approximation is 27.30.
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Suppose a skydiver must land on a target of three concentric circles. If the diameter of the center circle is 2 yards and the circles are spaced 1 yard apart, what is the probability that the skydiver will land in the red circle?
Find the following probability.
a. P(skydiver lands in the blue region)
The probability that the skydiver will land in the red circle is the area of the red circle divided by the area of the blue region the probability that the skydiver will land in the red circle is 1/5.
To find the probability that the skydiver will land in the red circle, we first need to determine the areas of the circles.
The diameter of the center circle is 2 yards, so the radius (half the diameter) is 1 yard.
Therefore, the area of the center circle is π * (1 yard)^2 = π square yards.
The next circle has a diameter of 2 + 2 * 1 = 4
yards, so the radius is 2 yards. The area of this circle is
π * (2 yards)^2 = 4π square yards.
The outermost circle has a diameter of 4 + 2 * 1 = 6
yards, so the radius is 3 yards. The area of this circle is π * (3 yards)^2 = 9π square yards.
To find the probability, we need to compare the area of the red circle (center circle) to the total area of the blue region (center and intermediate circles).
The area of the blue region is the sum of the areas of the center and intermediate circles: π square yards + 4π square yards = 5π square yards.
Therefore, the probability that the skydiver will land in the red circle is the area of the red circle divided by the area of the blue region:
P(skydiver lands in the blue region) = (π square yards) / (5π square yards) = 1/5.
So, the probability that the skydiver will land in the red circle is 1/5.
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Suppose {v1, v2, v3} is a linearly independent set of vectors in R3 and
let w = a1v1 + a2v2 + a3v3, with real numbers a1, a2, a3, be a linear
combination of these vectors. Prove the following statement: The
vectors w, v2, v3 are linearly independent if, and only if, a1 6= 0.
Hint: To show one implication, assume 0 = x1w+x2v2+x3v3 for some
numbers x1, x2, x3, and use that v1, v2, v3 are linearly independent to
derive that all xis must be zero.
1. If w, v2, v3 are linearly independent, then a1 ≠ 0:
Assume that w, v2, v3 are linearly independent. Suppose, for contradiction, that a1 = 0. Then we can express w as w = 0v1 + a2v2 + a3v3 = a2v2 + a3v3. Since v2 and v3 are linearly independent, we must have a2 = 0 and a3 = 0 for w to be linearly independent from v2 and v3.
However, this implies that w = 0, which contradicts the assumption that w is nonzero. Therefore, a1 must be nonzero.
2. If a1 ≠ 0, then w, v2, v3 are linearly independent:
Assume that a1 ≠ 0. We want to show that if x1w + x2v2 + x3v3 = 0, then x1 = x2 = x3 = 0. Substituting the expression for w, we have x1(a1v1) + x2v2 + x3v3 = 0. Since {v1, v2, v3} is linearly independent, the coefficients of v1, v2, and v3 must be zero. This gives us the following system of equations: x1a1 = 0, x2 = 0, and x3 = 0. Since a1 ≠ 0, the equation x1a1 = 0 implies that x1 = 0. Thus, x1 = x2 = x3 = 0, showing that the vectors are linearly independent.
Therefore, we have shown both implications, concluding that the vectors w, v2, v3 are linearly independent if and only if a1 ≠ 0.
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Find the slope and the y-intercept for the line with the
equation 2y+5x=-7
Therefore, the slope of the line is -5/2 and the y-intercept is -7/2.
To find the slope and y-intercept of the line with the equation 2y + 5x = -7, we need to rearrange the equation into the slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept.
Starting with the given equation:
2y + 5x = -7
We isolate y by subtracting 5x from both sides:
2y = -5x - 7
Divide both sides by 2 to solve for y:
y = (-5/2)x - 7/2
Comparing this equation with the slope-intercept form y = mx + b, we can see that the slope (m) is -5/2 and the y-intercept (b) is -7/2.
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Consider the initial value problem y ′
=x 2
,y(3)=5. Use Euler's method with a step size of 0.6, and starting at 3 , to find the approximate value for the solution to the initial value problem for x=4.2. Round your answer to three decimal places, but do not round any numbers until then.
The approximate solution for the initial value problem for x = 4.2 is 747.57 (rounded to three decimal places).
The basic idea behind Euler's method is to approximate the solution of an ODE by using small time steps and approximating the derivative of the function at each time step. The method is based on the tangent line approximation.
The Euler's method is given by;
y1 = y0 + hf(x0, y0)
By substituting the given values in the above equation,
y1 = 5 + 0.6 (3)² = 14.7
Now, use a table to calculate the approximate solution at x = 4.2;x yn0 53.000.60 141.799.20 346.1314.40 747.57 (approximate solution at x = 4.2)
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Let X and Y be sets, let R be a partial order on X, and S be a partial order on Y. Show that R∗S:={((x,y),(x′,y′)∈X×Y∣(x,x′)∈R,(y,y′)∈S} defines a partial order on X×Y. In other words, (x,y)(R∗S)(x′,y′) if and only if both xRx′ and ySy′. 4. In the context of the previous question, if R and S are total, must R∗S be total?
1. R∗S defines a partial order on X×Y because it satisfies reflexivity, antisymmetry, and transitivity.
2. R∗S is not necessarily total even if R and S are total. The totality of R and S only guarantees comparability within their respective sets, but not between elements in X and Y under R∗S.
To show that R∗S defines a partial order on X×Y, we need to demonstrate that it satisfies three properties: reflexivity, antisymmetry, and transitivity.
1. Reflexivity:
For any (x, y) ∈ X×Y, we want to show that (x, y) (R∗S) (x, y). According to the definition of R∗S, this means we need to have both xRx and ySy. Since R and S are both partial orders, they satisfy reflexivity. Therefore, (x, y) (R∗S) (x, y) holds.
2. Antisymmetry:
Suppose (x, y) (R∗S) (x', y') and (x', y') (R∗S) (x, y). This implies that both xRx' and ySy' as well as x'Rx and y'Sy. By the antisymmetry property of R and S, we have x = x' and y = y'. Thus, (x, y) = (x', y'), which satisfies the antisymmetry property of a partial order.
3. Transitivity:
If (x, y) (R∗S) (x', y') and (x', y') (R∗S) (x'', y''), it means that xRx' and ySy', as well as x'Rx'' and y'Sy''. Since R and S are both partial orders, we have xRx'' and ySy''. Hence, (x, y) (R∗S) (x'', y''), satisfying the transitivity property.
Therefore, we have shown that R∗S defines a partial order on X×Y.
4. If R and S are total, must R∗S be total?
No, R∗S is not necessarily total even if R and S are total. The total order of R and S only guarantees that every pair of elements in X and Y, respectively, are comparable. However, it does not ensure that every pair in X×Y will be comparable under R∗S. For R∗S to be total, every pair ((x, y), (x', y')) in X×Y would need to satisfy either (x, y)(R∗S)(x', y') or (x', y')(R∗S)(x, y). This is not guaranteed solely based on the totality of R and S, as the ordering relation may not hold between elements in different subsets of X×Y.
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d. If \( f \) has a removable discontinuity at \( x=5 \) and \( \lim _{x \rightarrow 5^{-}} f(x)=2 \), then \( f(5)= \) i. 2 ii. 5 iii. \( \infty \) iv. The limit does not exist v. Cannot be determine
The statement is true because for any function with a removable discontinuity, the value at the point is always equal to the limit from both sides.
Therefore, if \( f \) has a removable discontinuity at \
( x=5 \) and \( \lim _{x \ rightar row 5^{-}} f(x)=2 \),
then \( f(5)=2\ 2It is given that \( f \) has a removable discontinuity at
\( x=5 \) and \
( \lim _{x \rightarrow 5^{-}} f(x)=2 \).
Removable Discontinuity is a kind of discontinuity in which the function is discontinuous at a point, but it can be fixed by defining or redefining the function at that particular point.
Therefore, we can say that for any function with a removable discontinuity, the value at the point is always equal to the limit from both sides. Hence, we can say that if \( f \) has a removable discontinuity at \
( x=5 \) and \( \lim _{x \rightarrow 5^{-}} f(x)=2 \), then \( f(5)=2\).
Therefore, the correct option is i. 2.
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Find a plane containing the point (−3,−6,−4) and the line r (t)=<−5,5,5>+t<−7,−1,−1>
the equation of the plane containing the point (-3, -6, -4) and the line r(t) = <-5, 5, 5> + t<-7, -1, -1> is 7x + y - z = -4.
To find the equation of a plane, we need a point on the plane and a direction vector perpendicular to the plane.
Given the point (-3, -6, -4), we can use it as a point on the plane.
For the direction vector, we can take the direction vector of the given line, which is <-7, -1, -1>. Since any scalar multiple of a direction vector will still be perpendicular to the plane, we can choose to multiply this vector by any non-zero scalar. In this case, we'll use the scalar 1.
Now, we have a point on the plane (-3, -6, -4) and a direction vector <-7, -1, -1>.
Using the point-normal form of the equation of a plane, we can write the equation as follows:
7(x - (-3)) + (y - (-6)) - (z - (-4)) = 0
Simplifying, we get:
7x + y - z = -4
Therefore, the equation of the plane containing the point (-3, -6, -4) and the line r(t) = <-5, 5, 5> + t<-7, -1, -1> is 7x + y - z = -4.
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Two solutions to y ′′
−4y ′
+3y=0 are y 1
=e t
,y 2
=e 3t
. a) Find the Wronskian. W= b) Find the solution satisfying the initial conditions y(0)=2,y ′
(0)=16 y=
The Wronskian is given by [tex]W(e^t, e^3^t) = 2e^4^t.[/tex] The solution satisfying the initial conditions y(0) = 2, y′(0) = 16 is [tex]y(t) = 2e^t.[/tex]
We are to find the Wronskian and a solution to y ″ − 4y′ + 3y = 0. Here are the steps to solve this problem:
Step 1: We are to find the Wronskian. The formula for the Wronskian is given by:
W(y1, y2) = y1y′2 − y2y′1.
W(e^t, e^3^t) = e^t(e^3^t)′ − e^3t(e^t)′
W(e^t, e^3t) = e^t(3e^3t) − e^3t(e^t)
W(e^t, e^3t) = 2e^4t
W(e^t, e^3t) = 2e^4t
We are to find the solution satisfying the initial conditions y(0) = 2, y′(0) = 16.
The general solution to y″ − 4y′ + 3y = 0 is given by y(t) = c1e^t + c2e^3t.
Differentiating the equation gives:y′(t) = c1e^t + 3c2e^3t
Plugging in y(0) = 2 gives:2 = c1 + c2
Plugging in y′(0) = 16 gives:16 = c1 + 3c2
Solving these equations gives:
c1 = 2c2 = 0
We can now solve for y(t) by plugging in the values for c1 and c2 into y(t) = c1e^t + c2e^3t.
y(t) = 2e^t
The Wronskian is given by W(e^t, e^3t) = 2e^4t. The solution satisfying the initial conditions y(0) = 2, y′(0) = 16 is y(t) = 2e^t.
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Assume the real rate of interest is 4.00% and the inflation rate is 6.00%. What is the value today of receiving 11,713.00 in 14.00 years?
The present value of receiving $11,713.00 in 14.00 years, considering a 4.00% real rate of interest and a 6.00% inflation rate, is approximately $6,620.33.
To find the present value, we use the formula for present value with inflation: PV = FV /[tex](1+r-i)^{n}[/tex] where PV is the present value, FV is the future value, r is the real rate of interest, i is the inflation rate, and n is the number of years.
Substituting the given values into the formula:
PV = 11,713.00 / [tex](1+0.04-0.06) ^{14}[/tex]
PV = 11,713.00 / [tex](1-0.02)^{14}[/tex]
PV = 11,713.00 / [tex]0.98^{14}[/tex]
Using a calculator, we can compute the present value to be approximately $6,620.33.
Therefore, the present value of receiving $11,713.00 in 14.00 years, considering the given real rate of interest and inflation rate, is approximately $6,620.33.
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1) Find x+y, if: 2x+3y=8 and
3x+5y=13
a. 1.5
b. 2
c. 2.5
d. 3
e. 3.5
2) Find x+y, if: 5x+8y=67 and 2x-y=31
a. 98
b. 46
c. 21
d. 14
e. 7
(1) Therefore, x + y = 1 + 2 = 3. The correct answer is (d) 3. (2)Therefore, x + y = 15 + (-1) = 14. The correct answer is (d) 14.
1. To find x + y, we can solve the system of equations:
2x + 3y = 8 ...(1)
3x + 5y = 13 ...(2)
We can multiply equation (1) by 3 and equation (2) by 2 to eliminate x:
6x + 9y = 24 ...(3)
6x + 10y = 26 ...(4)
Subtracting equation (3) from equation (4), we get:
y = 2
Substituting this value of y into equation (1), we can solve for x:
2x + 3(2) = 8
2x + 6 = 8
2x = 2
x = 1
Therefore, x + y = 1 + 2 = 3. The correct answer is (d) 3.
2. we have the system of equations:
5x + 8y = 67 ...(5)
2x - y = 31 ...(6)
We can solve equation (6) for y:
y = 2x - 31
Substituting this value of y into equation (5), we have:
5x + 8(2x - 31) = 67
5x + 16x - 248 = 67
21x - 248 = 67
21x = 315
x = 15
Substituting x = 15 into equation (6), we can solve for y:
2(15) - y = 31
30 - y = 31
y = -1
Therefore, x + y = 15 + (-1) = 14. The correct answer is (d) 14.
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Find the length of the line segment from A(0,2) to B(2,4)
The length of the line segment from point [tex]A(0, 2)[/tex] to point [tex]B(2, 4) is \(2 \cdot \sqrt{{2}}\)[/tex] units.
To find the length of the line segment from point A(0, 2) to point B(2, 4), we can use the distance formula. The distance formula calculates the length of a line segment between two points in a coordinate plane.
The distance formula is given by:
[tex]\(d = \sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}}\)[/tex]
Let's substitute the coordinates of point A and point B into the formula:
[tex]\(d = \sqrt{{(2 - 0)^2 + (4 - 2)^2}}\)[/tex]
Simplifying the expression:
\(d = \sqrt{{2^2 + 2^2}}\)
\(d = \sqrt{{4 + 4}}\)
\(d = \sqrt{{8}}\)
To simplify further, we can write \(8\) as \(4 \cdot 2\):
\(d = \sqrt{{4 \cdot 2}}\)
Using the property of square roots, we can split the square root:
\(d = \sqrt{{4}} \cdot \sqrt{{2}}\)
\(d = 2 \cdot \sqrt{{2}}\)
Therefore, the length of the line segment from point A(0, 2) to point B(2, 4) is \(2 \cdot \sqrt{{2}}\) units.
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find the volume of the solid obtained by rotating the region
bounded by y=x and y= sqrt(x) about the line x=2
Find the volume of the solid oblained by rotating the region bounded by \( y=x \) and \( y=\sqrt{x} \) about the line \( x=2 \). Volume =
The volume of the solid obtained by rotating the region bounded by \[tex](y=x\) and \(y=\sqrt{x}\)[/tex] about the line [tex]\(x=2\) is \(\frac{-2}{3}\pi\) or \(\frac{2}{3}\pi\)[/tex] in absolute value.
To find the volume of the solid obtained by rotating the region bounded by \(y=x\) and \(y=\sqrt{x}\) about the line \(x=2\), we can use the method of cylindrical shells.
The cylindrical shells are formed by taking thin horizontal strips of the region and rotating them around the axis of rotation. The height of each shell is the difference between the \(x\) values of the curves, which is \(x-\sqrt{x}\). The radius of each shell is the distance from the axis of rotation, which is \(2-x\). The thickness of each shell is denoted by \(dx\).
The volume of each cylindrical shell is given by[tex]\(2\pi \cdot (2-x) \cdot (x-\sqrt{x}) \cdot dx\)[/tex].
To find the total volume, we integrate this expression over the interval where the two curves intersect, which is from \(x=0\) to \(x=1\). Therefore, the volume can be calculated as follows:
\[V = \int_{0}^{1} 2\pi \cdot (2-x) \cdot (x-\sqrt{x}) \, dx\]
We can simplify the integrand by expanding it:
\[V = \int_{0}^{1} 2\pi \cdot (2x-x^2-2\sqrt{x}+x\sqrt{x}) \, dx\]
Simplifying further:
\[V = \int_{0}^{1} 2\pi \cdot (x^2+x\sqrt{x}-2x-2\sqrt{x}) \, dx\]
Integrating term by term:
\[V = \pi \cdot \left(\frac{x^3}{3}+\frac{2x^{\frac{3}{2}}}{3}-x^2-2x\sqrt{x}\right) \Bigg|_{0}^{1}\]
Evaluating the definite integral:
\[V = \pi \cdot \left(\frac{1}{3}+\frac{2}{3}-1-2\right)\]
Simplifying:
\[V = \pi \cdot \left(\frac{1}{3}-1\right)\]
\[V = \pi \cdot \left(\frac{-2}{3}\right)\]
Therefore, the volume of the solid obtained by rotating the region bounded by \(y=x\) and \(y=\sqrt{x}\) about the line \(x=2\) is \(\frac{-2}{3}\pi\) or \(\frac{2}{3}\pi\) in absolute value.
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The rules for a race require that all runners start at $A$, touch any part of the 1200-meter wall, and stop at $B$. What is the number of meters in the minimum distance a participant must run
The number of meters in the minimum distance a participant must run is 800 meters.
The minimum distance a participant must run in this race can be calculated by finding the length of the straight line segment between points A and B. This can be done using the Pythagorean theorem.
Given that the participant must touch any part of the 1200-meter wall, we can assume that the shortest distance between points A and B is a straight line.
Using the Pythagorean theorem, the length of the straight line segment can be found by taking the square root of the sum of the squares of the lengths of the two legs. In this case, the two legs are the distance from point A to the wall and the distance from the wall to point B.
Let's assume that the distance from point A to the wall is x meters. Then the distance from the wall to point B would also be x meters, since the participant must stop at point B.
Applying the Pythagorean theorem, we have:
x^2 + 1200^2 = (2x)^2
Simplifying this equation, we get:
x^2 + 1200^2 = 4x^2
Rearranging and combining like terms, we have:
3x^2 = 1200^2
Dividing both sides by 3, we get:
x^2 = 400^2
Taking the square root of both sides, we get:
x = 400
Therefore, the distance from point A to the wall (and from the wall to point B) is 400 meters.
Since the participant must run from point A to the wall and from the wall to point B, the total distance they must run is twice the distance from point A to the wall.
Therefore, the minimum distance a participant must run is:
2 * 400 = 800 meters.
So, the number of meters in the minimum distance a participant must run is 800 meters.
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The minimum distance a participant must run in the race, we need to consider the path that covers all the required points. First, the participant starts at point A. Then, they must touch any part of the 1200-meter wall before reaching point B. The number of meters in the minimum distance a participant must run in this race is 1200 meters.
To minimize the distance, the participant should take the shortest path possible from A to B while still touching the wall.
Since the wall is a straight line, the shortest path would be a straight line as well. Thus, the participant should run directly from point A to the wall, touch it, and continue running in a straight line to point B.
This means the participant would cover a distance equal to the length of the straight line segment from A to B, plus the length of the wall they touched.
Therefore, the minimum distance a participant must run is the sum of the distance from A to B and the length of the wall, which is 1200 meters.
In conclusion, the number of meters in the minimum distance a participant must run in this race is 1200 meters.
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Find the absolute extrema of f(x,y,z)=x^2−y^2 subject to the constraint x^2+2y^2+3z^2=1. Show all points you considered in the process. (You may assume that an absolute max and an absolute min exist)
We are given that the function f(x,y,z)=x²−y² has the constraint x²+2y²+3z²=1 and we have to find the absolute extrema of the function under this constraint.So, let's solve the problem:
First, we find the gradient of the function f(x,y,z):
[tex]gradf(x,y,z)= [∂f/∂x, ∂f/∂y, ∂f/∂z] = [2x, -2y, 0][/tex]
Now we will find the Lagrange multiplier by the following equation:
[tex]gradf(x,y,z) = λ * gradg(x,y,z)[/tex] where [tex]g(x,y,z)= x²+2y²+3z² -1gradg(x,y,z)= [2x, 4y, 6z][/tex]
Using these, we get:
[tex]2x = λ * 2x-2y = λ * 4y0 = λ * 6z[/tex]
From the first equation, either x=0 or λ=1. If x=0, then we have y=0 and z= ±1/√3, which corresponds to the values of x, y, and z where the value of f(x,y,z) is ±1/3. If λ=1,
then we have x=y/2. Using the equation for g(x,y,z), we get:
y²+3z²=1/2, y=z√2/3. This corresponds to the value of x, y, and z where the value of f(x,y,z) is 1/3
Thus, the absolute extrema of f(x,y,z)=x²−y² under the constraint x²+2y²+3z²=1 are -1/3 and 1/3. The corresponding points are (0,0,±1/√3) and (1/√6, ±1/√6, ±1/√3) respectively.
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in a family of nine children, what is the probability of having either three boys and six girls or four boys and five girls?
The probability of having either three boys and six girls or four boys and five girls in a family of nine children is 210/512.
To calculate this probability, we can use the following formula:
Probability = (Number of desired outcomes)/(Total number of possible outcomes)
The total number of possible outcomes is 2^9 = 512, because there are 2 possible genders for each child, and there are 9 children in total.
The number of desired outcomes is the number of ways to have either 3 boys and 6 girls or 4 boys and 5 girls. There are 84 ways to have 3 boys and 6 girls, and 126 ways to have 4 boys and 5 girls.
Therefore, the probability of having either 3 boys and 6 girls or 4 boys and 5 girls in a family of nine children is:
Probability = (84 + 126) / 512 = 210 / 512
This is a relatively rare event, with a probability of only about 4%.
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Writing Equations Parallel & Perpendicular Lines.
1. Write the slope-intercept form of the equation of the line described. Through: (2,2), parallel y= x+4
2. Through: (4,3), Parallel to x=0.
3.Through: (1,-5), Perpendicular to Y=1/8x + 2
Equation of the line described: y = x + 4
Slope of given line y = x + 4 is 1
Therefore, slope of parallel line is also 1
Using the point-slope form of the equation of a line,
we have y - y1 = m(x - x1),
where (x1, y1) = (2, 2)
Substituting the values, we get
y - 2 = 1(x - 2)
Simplifying the equation, we get
y = x - 1
Therefore, slope-intercept form of the equation of the line is
y = x - 12.
Equation of the line described:
x = 0
Since line is parallel to the y-axis, slope of the line is undefined
Therefore, the equation of the line is x = 4.3.
Equation of the line described:
y = (1/8)x + 2
Slope of given line y = (1/8)x + 2 is 1/8
Therefore, slope of perpendicular line is -8
Using the point-slope form of the equation of a line,
we have y - y1 = m(x - x1),
where (x1, y1) = (1, -5)
Substituting the values, we get
y - (-5) = -8(x - 1)
Simplifying the equation, we get y = -8x - 3
Therefore, slope-intercept form of the equation of the line is y = -8x - 3.
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5. Determine how many positive and negative zeroes the following polynomial function \( f(x)=2 x^{3}-4 x^{2}-10 x+20 \) may have. 6. From the above function of question- 05 , find all real zeroes.
These are the two real zeroes of the given polynomial function f(x) = 2x³- 4x² - 10x + 20.
To determine the number of positive and negative zeroes of the polynomial function f(x) = 2x³- 4x² - 10x + 20 , we need to examine the sign changes in the coefficients.
By counting the sign changes in the coefficients, we can determine the maximum number of positive and negative zeroes. However, this method does not guarantee the exact number of zeroes; it only provides an upper limit.
Let's write down the coefficients of the polynomial:
f(x) = 2x³- 4x² - 10x + 20
The sign changes in the coefficients are as follows:
From (2) to (-4), there is a sign change.
From (-4) to (-10), there is no sign change.
From (-10) to (20), there is a sign change.
So, based on the sign changes, the polynomial f(x) can have at most:
- 1 positive zero
- 1 or 3 negative zeroes
Now, let's find all the real zeroes of the polynomial function f(x) = 2x³- 4x² - 10x + 20.
To find the real zeroes, we can use methods like factoring, synthetic division, or numerical approximation techniques.
Using numerical approximation, we can find the real zeroes to be approximately:
x ≈ -1.847
x ≈ 1.847
These are the two real zeroes of the given polynomial function.
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1. Let A be a 3×7 matrix. Answer each of the following questions about A. If the solution cannot be determined with the given information, write CANNOT BE DETERMINED. (a) If the product Av is defined for column vector v, what is the size of v ? (b) If T is the linear transformation defined by T(x)=Ax, what is the domain of T ?
(a) The size of v is 7. (b) Since matrix A is a 3×7 matrix, it can multiply with a column vector of size 7. Therefore, the domain of T is the set of column vectors of size 7.
(a) If the product Av is defined fhttps://brainly.com/question/28180105or column vector v, the number of columns in matrix A must be equal to the number of rows in vector v. In this case, A is a 3×7 matrix, so v must be a column vector with 7 elements.
Therefore, the size of v is 7.
(b) The linear transformation T(x) = Ax is defined by multiplying matrix A with vector x. The domain of T is the set of all vectors x for which the transformation T(x) is defined.
Since matrix A is a 3×7 matrix, it can multiply with a column vector of size 7. Therefore, the domain of T is the set of column vectors of size 7.
In summary, the domain of the linear transformation T is the set of column vectors of size 7.
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Given f (x)=−16sin(4x) and f (0)=−6 and f(0)=−5. Find f( 3/π )=
Answer:
Step-by-step explanation:
To find the value of f(3/π), we need to substitute x = 3/π into the function f(x) = -16sin(4x).
f(3/π) = -16sin(4 * (3/π))
Since sin(π) = 0, we can simplify the expression further:
f(3/π) = -16sin(4 * (3/π)) = -16sin(12/π)
Now, we need to evaluate sin(12/π). Remember that sin(θ) = 0 when θ is an integer multiple of π. Since 12/π is not an integer multiple of π, we cannot simplify it further.
Therefore, the value of f(3/π) is -16sin(12/π), and we do not have enough information to determine its numerical value without additional calculations or approximations.
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you want to choose an srs of 20 of indiana’s 5341 voting precincts for special voting-fraud surveillance on election day. (a) to choose an srs, how many digits would you need to make up each of your labels for the 5341 precincts?
To make up each label for the 5,341 precincts, we would need four digits.
To choose a simple random sample (SRS) of 20 voting precincts from Indiana's 5,341 precincts, we need to assign labels to each precinct. The number of digits required for each label depends on the maximum number of precincts we have.
In this case, since we have 5,341 precincts, the maximum label we would need to assign is 5,341. To determine the number of digits needed, we count the number of digits in this maximum label.
The maximum label has four digits (5,341). Therefore, to make up each label for the 5,341 precincts, we would need four digits.
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the Cartesian product Z 2
⊗Z 5
of two sets of congruence classes, Z 2
and Z 5
, under operations ([k],[m])⊞([l],[n]):=([k+l],[m+n]) and ([k],[m])□([l],[n]):=([kl],[mn]) (a) Prove that the first distributive law holds true. (b) Hence prove that ≺Z 2
×Z 5
,⊞,□≻ is a ring. (c) Is it a commutative ring? Justify your answer.
The first distributive law holds true. All the ring axioms hold, it is a ring. The distributive law is not commutative in general, this ring is not commutative.
a) Let A, B, and C be subsets of a set X.
Distributive law states that: (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C).
Since the first distributive law requires the verification of the equality of two sets, we must demonstrate that:
(a, b)□((c, d)⊞(e, f)) ≡ ((a, b)□(c, d))⊞((a, b)□(e, f))
Therefore, we must evaluate the two sides separately.
We have:
(a, b)□((c, d)⊞(e, f)) = (a, b)□(c+e, d+f) = (ac + ae, bd + bf),((a, b)□(c, d))⊞((a, b)□(e, f)) = (ac, bd)⊞(ae, bf) = (ac + ae, bd + bf)
So, the first distributive law holds true.
b) Using the first distributive law from part a), we can demonstrate that ≺Z2×Z5,⊞,□≻ is a ring.
We must verify that the following properties hold for each pair of elements (x, y), (z, w) in Z2×Z5:
(i) Closure under ⊞: (x, y)⊞(z, w) ∈ Z2×Z5. This follows from the closure of Z2 and Z5 under addition.
(ii) Closure under □: (x, y)□(z, w) ∈ Z2×Z5. This follows from the closure of Z2 and Z5 under multiplication.
(iii) Associativity under ⊞: ((x, y)⊞(z, w))⊞(a, b) = (x, y)⊞((z, w)⊞(a, b)). Associativity of addition in Z2 and Z5 ensures that this property holds true.
(iv) Identity under ⊞: There exists an element (0, 0) ∈ Z2×Z5 such that (x, y)⊞(0, 0) = (x, y) for all (x, y) ∈ Z2×Z5. The additive identity elements in Z2 and Z5 make this true.
(v) Inverse under ⊞: For any element (x, y) ∈ Z2×Z5, there exists an element (z, w) ∈ Z2×Z5 such that (x, y)⊞(z, w) = (0, 0). This follows from the closure of Z2 and Z5 under addition, and the existence of additive inverses.
(vi) Associativity under □: ((x, y)□(z, w))□(a, b) = (x, y)□((z, w)□(a, b)). Associativity of multiplication in Z2 and Z5 ensures that this property holds true.
(vii) Distributive law: (x, y)□((z, w)⊞(a, b)) = (x, y)□(z, w)⊞(x, y)□(a, b). This property is verified in part a).
(viii) Commutativity under ⊞: (x, y)⊞(z, w) = (z, w)⊞(x, y). Commutativity of addition in Z2 and Z5 ensures that this property holds true.
(ix) Commutativity under □: (x, y)□(z, w) = (z, w)□(x, y). Commutativity of multiplication in Z2 and Z5 ensures that this property holds true.
Since all the ring axioms hold, it is a ring.
c) Since commutativity under ⊞ and □ is required to establish a commutative ring, and the distributive law is not commutative in general, this ring is not commutative.
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