The number of stores in the tower, and y represents the number of squares. Press “Graph” to view the graph. The graph is given below:Graph of the function rule p = 4s + 1.
Given that the function rule that represents the relationship between the number of stores in the tower, s, and the number of squares, p is p = 4s + 1. To graph the given function, follow the steps below:
1: Select the data that you want to plot.
2: Enter the data into the graphing calculator.
3: Choose a graph type. Here, we can choose scatter plot as we are plotting data points.
4: Press the “Graph” button to view the graph.
5: To graph the function rule, select the “y=” button and enter the equation as y = 4x + 1.
Here, x represents the number of stores in the tower, and y represents the number of squares. Press “Graph” to view the graph. The graph is given below: Graph of the function rule p = 4s + 1.
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Find out the decimal equivalent of 1011 if the given number is an/a a) Unsigned number b) Negative number using Signed magnitude c) Negative number using I's complement d) Negative number using 2's complement Consider a 4-bit system including sign (1 bit represents sign and 3 bits represent magnitude) for representing signed numbers.
a) The decimal equivalent of the unsigned number 1011 is 11.
b) The decimal equivalent of the negative number using signed magnitude is -11.
c) The decimal equivalent of the negative number using 1's complement is -4.
d) The decimal equivalent of the negative number using 2's complement is -5.
a) To convert the unsigned binary number 1011 to decimal, we simply calculate the value of each bit position. In this case, the decimal equivalent is 1*(2^3) + 0*(2^2) + 1*(2^1) + 1*(2^0) = 8 + 0 + 2 + 1 = 11.
b) In signed magnitude representation, the leftmost bit represents the sign of the number. In this case, the leftmost bit is 1, indicating a negative number. To obtain the magnitude, we convert the remaining three bits (011) to decimal, which is 3. Therefore, the decimal equivalent of -1011 in signed magnitude is -11.
c) In 1's complement representation, negative numbers are obtained by taking the bitwise complement of the magnitude bits. In this case, the magnitude bits are 011, and the 1's complement of 011 is 100. Therefore, the decimal equivalent of -1011 in 1's complement is -4.
d) In 2's complement representation, negative numbers are obtained by taking the 2's complement of the magnitude bits. To find the 2's complement, we first take the 1's complement of the magnitude (011) which gives us 100. Then we add 1 to the 1's complement: 100 + 1 = 101. Therefore, the decimal equivalent of -1011 in 2's complement is -5.
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If the value of a $29,575car decreases by 25% each year due to depreciation, how much will the car be worth after 14 years? Round your answer to the nearest dollar (whole number). Do not enter the dollar sign. For example, if the answer is $5500, type 5500 .
The value worth of the car after 14 years will be $38,259 (Rounded to the nearest dollar).
Given information:Value of a $29,575 car decreases by 25% each year due to depreciation. We need to find how much will the car be worth after 14 years using the above information.Therefore,To find out how much the car will worth after 14 years, we can use the below formula:P = (1-r/n)^nt × P₀Where,P₀ = Initial value of the carP = Worth of the car after n yearsr = Annual depreciation rate (0.25) or 25%t = Number of years elapsedWe can substitute the given values in the above formula:P = (1-0.25/1)^14 × 29,575= (0.75)^14 × 29,575= 1294.632 × 29,575= 38,258.97 (Rounded to the nearest whole number)= $38,259 (Rounded to the nearest dollar)Hence, the worth of the car after 14 years will be $38,259 (Rounded to the nearest dollar).
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The point k lies on the segment jk find the coordinates of k si that jk is 1/5 of jl
To find the coordinates of point K on the segment JL such that JK is 3/7 of JL, we calculate the change in x and y coordinates from J to L, multiply them by 3/7, and add them to the coordinates of J. This gives us the coordinates of point K as (-9, 5).
To find the coordinates of point K, we need to determine a point that lies on the segment JL and is 3/7 of the distance from J to L.
Step 1: Find the difference between the x-coordinates of J and L:
Δx = Lx - Jx = 3 - (-18) = 21
Step 2: Find the difference between the y-coordinates of J and L:
Δy = Ly - Jy = (-11) - 17 = -28
Step 3: Multiply the differences by 3/7 to find the change in x and y coordinates from J to K:
Δx' = (3/7) * Δx = (3/7) * 21 = 9
Δy' = (3/7) * Δy = (3/7) * (-28) = -12
Step 4: Add the change in x and y coordinates to the coordinates of J to find the coordinates of K:
Kx = Jx + Δx' = -18 + 9 = -9
Ky = Jy + Δy' = 17 + (-12) = 5
Therefore, the coordinates of point K are (-9, 5).
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The complete question is:
The point K lies on the segment JL.
Find the coordinates of K so that JK is 3/7 of JL.
J= (-18, 17)
K=(?,?)
L= (3,-11)
Find the coordinates of K.
. Determine the standard equation of the ellipse using the stated information.
Foci at (8,−1) and (−2,−1); length of the major axis is twelve units
The equation of the ellipse in standard form is _____.
b. Determine the standard equation of the ellipse using the stated information.
Vertices at (−5,12) and (−5,2); length of the minor axis is 8 units.
The standard form of the equation of this ellipse is _____.
c. Determine the standard equation of the ellipse using the stated information.
Center at (−4,1); vertex at (−4,10); focus at (−4,9)
The equation of the ellipse in standard form is ____.
a. The standard equation of the ellipse with foci at (8, -1) and (-2, -1), and a length of the major axis of 12 units is: ((x - 5)² / 6²) + ((y + 1)² / b²) = 1.
b. The standard equation of the ellipse with vertices at (-5, 12) and (-5, 2), and a length of the minor axis of 8 units is: ((x + 5)² / a²) + ((y - 7)² / 4²) = 1.
c. The standard equation of the ellipse with a center at (-4, 1), a vertex at (-4, 10), and a focus at (-4, 9) is: ((x + 4)² / b²) + ((y - 1)² / 9²) = 1.
a. To determine the standard equation of the ellipse with foci at (8, -1) and (-2, -1), and a length of the major axis of 12 units, we can start by finding the distance between the foci, which is equal to the length of the major axis.
Distance between the foci = 12 units
The distance between two points (x₁, y₁) and (x₂, y₂) is given by the formula:
√((x₂ - x₁)² + (y₂ - y₁)²)
Using this formula, we can calculate the distance between the foci:
√((8 - (-2))² + (-1 - (-1))²) = √(10²) = 10 units
Since the distance between the foci is equal to the length of the major axis, we can conclude that the major axis of the ellipse lies along the x-axis.
The center of the ellipse is the midpoint between the foci, which is (5, -1).
The equation of an ellipse with a center at (h, k), a major axis of length 2a along the x-axis, and a minor axis of length 2b along the y-axis is:
((x - h)² / a²) + ((y - k)² / b²) = 1
In this case, the center is (5, -1) and the major axis is 12 units, so a = 12/2 = 6.
Therefore, the equation of the ellipse in standard form is:
((x - 5)² / 6²) + ((y + 1)² / b²) = 1
b. To determine the standard equation of the ellipse with vertices at (-5, 12) and (-5, 2), and a length of the minor axis of 8 units, we can start by finding the distance between the vertices, which is equal to the length of the minor axis.
Distance between the vertices = 8 units
The distance between two points (x₁, y₁) and (x₂, y₂) is given by the formula:
√((x₂ - x₁)² + (y₂ - y₁)²)
Using this formula, we can calculate the distance between the vertices:
√((-5 - (-5))² + (12 - 2)²) = √(0² + 10²) = 10 units
Since the distance between the vertices is equal to the length of the minor axis, we can conclude that the minor axis of the ellipse lies along the y-axis.
The center of the ellipse is the midpoint between the vertices, which is (-5, 7).
The equation of an ellipse with a center at (h, k), a major axis of length 2a along the x-axis, and a minor axis of length 2b along the y-axis is:
((x - h)² / a²) + ((y - k)² / b²) = 1
In this case, the center is (-5, 7) and the minor axis is 8 units, so b = 8/2 = 4.
Therefore, the equation of the ellipse in standard form is:
((x + 5)² / a²) + ((y - 7)² / 4²) = 1
c. To determine the standard equation of the ellipse with a center at (-4, 1), a vertex at (-4, 10), and a focus at (-4, 9), we can observe that the major axis of the ellipse is vertical, along the y-axis.
The distance between the center and the vertex gives us the value of a, which is the distance from the center to either focus.
a = 10 - 1 = 9 units
The distance between the center and the focus gives us the value of c, which is the distance from the center to either focus.
c = 9 - 1 = 8 units
The equation of an ellipse with a center at (h, k), a major axis of length 2a along the y-axis, and a distance c from the center to either focus is:
((x - h)² / b²) + ((y - k)² / a²) = 1
In this case, the center is (-4, 1), so h = -4 and k = 1.
Therefore, the equation of the ellipse in standard form is:
((x + 4)² / b²) + ((y - 1)² / 9²) = 1
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Solve each inequality. (Lesson 0-6) -14 n ≥ 42
To solve the inequality [tex]-14n ≥ 42[/tex], we need to isolate the variable n. Now, we know that the solution to the inequality [tex]-14n ≥ 42[/tex] is [tex]n ≤ -3.[/tex]
To solve the inequality -14n ≥ 42, we need to isolate the variable n.
First, divide both sides of the inequality by -14.
Remember, when dividing or multiplying both sides of an inequality by a negative number, you need to reverse the inequality symbol.
So, [tex]-14n / -14 ≤ 42 / -14[/tex]
Simplifying this, we get n ≤ -3.
Therefore, the solution to the inequality -14n ≥ 42 is n ≤ -3.
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Since 56 is greater than or equal to 42, the inequality is true.
To solve the inequality -14n ≥ 42, we need to isolate the variable n.
First, let's divide both sides of the inequality by -14. Remember, when dividing or multiplying an inequality by a negative number, we need to reverse the inequality symbol.
-14n ≥ 42
Divide both sides by -14:
n ≤ -3
So the solution to the inequality -14n ≥ 42 is n ≤ -3.
This means that any value of n that is less than or equal to -3 will satisfy the inequality. To verify this, you can substitute a value less than or equal to -3 into the original inequality and see if it holds true. For example, if we substitute -4 for n, we get:
-14(-4) ≥ 42
56 ≥ 42
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Evaluate f(x)=∑[infinity] to n=0 3^−n x^n at x=−1.
we cannot evaluate f(-1) as the series does not converge for x = -1.
To evaluate the given series, we substitute the value of x = -1 into the expression:
f(x) = ∑[n=0 to ∞] (3⁻ⁿ xⁿ)
Substituting x = -1:
f(-1) = ∑[n=0 to ∞] (3⁻ⁿ (-1)ⁿ)
Now, let's examine the behavior of the series. Notice that for n ≥ 1, the terms alternate between positive and negative as (-1)ⁿ changes sign. Therefore, the series does not converge for x = -1.
The series is a geometric series with a common ratio of (x/3). For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, when x = -1, the common ratio is (-1/3), which has an absolute value greater than 1, making the series divergent.
Therefore, we cannot evaluate f(-1) as the series does not converge for x = -1.
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6.4 q5 please solve a-b
\( y=x, y=x^{\frac{1}{9}} \) in the first quadrant; revolved about the \( x \)-axis choice. A. \( d x+\int \quad d x \) (Type exact answers.) B. \( \int \) dx (Type exact answers.)
The volume of the solid formed is \(\frac{2\pi}{75}\) cubic units.Answer: \(\boxed{\frac{2\pi}{75}}\)
Given: Two curves, \(y=x\) and \(y=x^{\frac19}\), in the first quadrant; revolved about the x-axis.We need to find the volume of the solid formed.Let's make the following observations regarding the curves:\(y=x\) intersects the x-axis at (0,0) and y-axis at (1,1)\(y=x^{\frac19}\) intersects the x-axis at (0,0) and y-axis at (1,1)Note that these points of intersection are the limits of integration.
Also, the curves intersect at (1,1).We can integrate by disks/washers method:\[\begin{align} V &=\pi \int_{0}^{1}[\left(r_{out}\right)^{2}-\left(r_{in}\right)^{2}] d x \\ &=\pi \int_{0}^{1}[(x)^{2}-(x^{1 / 9})^{2}] d x \end{align}\]We know that the volume of the solid of revolution is given by:\[V = \pi \int_a^b f^2(x) dx\]where the function \(f(x)\) is the distance from the curve to the axis of revolution.
To find the limits of integration, we will equate the two curves:$$x=x^{\frac19} \implies x^{\frac98}-x=0 \implies x(x^{\frac78}-1)=0$$So, the curves intersect at the origin and at (1,1).Therefore,\[\begin{aligned} V &=\pi \int_{0}^{1}\left[(x)^{2}-\left(x^{\frac{1}{9}}\right)^{2}\right] d x \\ &=\pi \int_{0}^{1}\left(x^{\frac{16}{9}}-x^{2}\right) d x \\ &=\pi\left[\frac{9}{25} x^{\frac{25}{9}}-\frac{1}{3} x^{3}\right]_{0}^{1} \\ &=\pi\left(\frac{9}{25}-\frac{1}{3}\right) \\ &=\boxed{\frac{2 \pi}{75}} \end{aligned}\]Therefore, the volume of the solid formed is \(\frac{2\pi}{75}\) cubic units.Answer: \(\boxed{\frac{2\pi}{75}}\)
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Question 1: [2 Mark] Find all eigen values and the associated eigenvectors for each of the following matrices a) A=[ 9
2
−10
0
] b) B= ⎣
⎡
1
−2
−6
0
3
6
0
−1
−1
⎦
⎤
a) The eigenvalues of matrix A are λ₁ = 92 and λ₂ = -100, with corresponding eigenvectors v₁ = [1, 1]ᵀ and v₂ = [1, -1]ᵀ.
b) The eigenvalues of matrix B are λ₁ = -2, λ₂ = -1, and λ₃ = -3, with corresponding eigenvectors v₁ = [2, 1, 0]ᵀ, v₂ = [1, 0, -1]ᵀ, and v₃ = [1, 1, 1]ᵀ.
To find the eigenvalues and eigenvectors of a given matrix, we need to solve the characteristic equation det(A - λI) = 0, where A is the matrix, λ represents the eigenvalues, and I is the identity matrix.
For matrix A, we have A = [92, -100]. Subtracting λ times the identity matrix of size 2 from A, we get the matrix A
- λI = [92-λ, -100; -100, -100-λ].
Calculating the determinant of A - λI and setting it equal to zero, we have (92-λ)(-100-λ) - (-100)(-100) = λ² - 8λ - 1800 = 0.
Solving this quadratic equation, we find the eigenvalues
λ₁ = 92 and λ₂ = -100.
To find the eigenvectors, we substitute each eigenvalue back into the equation (A - λI)v = 0 and solve for v.
For λ₁ = 92, we have
(A - 92I)v₁ = 0,
which simplifies to
[0, -100; -100, -192]v₁ = 0.
Solving this system of equations, we find
v₁ = [1, 1]ᵀ.
For λ₂ = -100, we have
(A - (-100)I)v₂ = 0,
which simplifies to
[192, -100; -100, 0]v₂ = 0.
Solving this system of equations, we find
v₂ = [1, -1]ᵀ.
For matrix B, we follow the same steps. Subtracting λ times the identity matrix of size 3 from B, we get the matrix B - λI. The characteristic equation becomes det(B - λI) = 0. Solving this equation, we find the eigenvalues λ₁ = -2, λ₂ = -1, and λ₃ = -3.
Substituting each eigenvalue back into the equation (B - λI)v = 0, we solve for the corresponding eigenvectors. For λ₁ = -2, we have (B - (-2)I)v₁ = 0, which simplifies to [3, -2, -6; 0, 3, 6; 0, 0, 1]v₁ = 0. Solving this system of equations, we find v₁ = [2, 1, 0]ᵀ.
For λ₂ = -1, we have (B - (-1)I)v₂ = 0, which simplifies to [2, -2, -6; 0, 2, 6; 0, 0, 0]v₂ = 0. Solving this system of equations, we find v₂ = [1, 0, -1]ᵀ.
For λ₃ = -3
we have (B - (-3)I)v₃ = 0, which simplifies to
[4, -2, -6; 0, 4, 6; 0, 0, 2]v₃ = 0
Solving this system of equations, we find
v₃ = [1, 1, 1]ᵀ.
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Using the zscore tables and the zscores you calculated above for Firms A and B, determine the probability that the stock price for Firm A or Firm B will fall below a penny.
NOTE: Please state your answer as a percent (e.g., X.XX%). Be sure to describe how you determined this combined probability in the space provided below.
Firm A z-score = -2.74
Firm B z-score = -2.21
The combined probability that the stock price for Firm A or Firm B will fall below a penny is approximately 0.29%.
To determine the combined probability, we can use the z-score tables. The z-score represents the number of standard deviations a data point is from the mean. In this case, the z-score for Firm A is -2.74, and the z-score for Firm B is -2.21.
To find the probability that the stock price falls below a penny, we need to find the area under the normal distribution curve to the left of a z-score of -2.74 for Firm A and the area to the left of a z-score of -2.21 for Firm B.
Using the z-score table, we can find that the area to the left of -2.74 is approximately 0.0033 or 0.33%. Similarly, the area to the left of -2.21 is approximately 0.0139 or 1.39%.
To determine the combined probability, we subtract the individual probabilities from 1 (since we want the probability of the stock price falling below a penny) and then multiply them together. So, the combined probability is (1 - 0.0033) * (1 - 0.0139) ≈ 0.9967 * 0.9861 ≈ 0.9869 or 0.9869%.
Therefore, the combined probability that the stock price for Firm A or Firm B will fall below a penny is approximately 0.29%.
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4.7.4
Please help with all parts thank you
Find the local maxima, local minima, and saddle points, if any, for the function \( z=2 x^{2}+2 y^{2}-12 x+8 y+2 \). (Use symbolic notation and fractions where needed. Give your answer as point coordi
The function[tex]\(z = 2x^2 + 2y^2 - 12x + 8y + 2\)[/tex] has a local minimum at the point (3, -2). There are no local maxima or saddle points for this function.
To find the critical points, we differentiate the function with respect to x and y. Taking the partial derivative with respect to x, we get:
[tex]\(\frac{{\partial z}}{{\partial x}} = 4x - 12\)[/tex]
Setting this derivative to zero, we solve for x:
[tex]\(4x - 12 = 0 \Rightarrow x = 3\)[/tex]
Taking the partial derivative with respect to y, we get:
[tex]\(\frac{{\partial z}}{{\partial y}} = 4y + 8\)[/tex]
Setting this derivative to zero, we solve for y:
[tex]\(4y + 8 = 0 \Rightarrow y = -2\)[/tex]
So the critical point is (3, -2).
Next, we analyze the second partial derivatives to determine the nature of this critical point. Taking the second partial derivative with respect to x, we get:
[tex]\(\frac{{\partial^2 z}}{{\partial x^2}} = 4\)[/tex]
Taking the second partial derivative with respect to y, we get:
[tex]\(\frac{{\partial^2 z}}{{\partial y^2}} = 4\)[/tex]
Taking the second partial derivative with respect to x and y, we get:
[tex]\(\frac{{\partial^2 z}}{{\partial x \partial y}} = 0\)[/tex]
Since both second partial derivatives are positive (4 and 4), and the determinant of the Hessian matrix [tex](\(4 \times 4 - 0^2\))[/tex] is positive, the critical point (3, -2) corresponds to a local minimum.
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from 1990 to 2001,german currency included coins called pfennigs, worth 1 pfennig each, and groschen, worth 10 pfennigs each. which equation represents the number of pfennig coins, p, and groschen coins, g, that have a combined value of 85 pfennigs?
The equation "p + 10g = 85" represents the connection between the number of pfennig coins (p) and groschen coins (g) needed to reach a total value of 85 pfennigs. Option B.
Let's set up the equations to represent the number of pfennig coins (p) and groschen coins (g) that have a combined value of 85 pfennigs.
First, let's establish the values of the coins:
1 pfennig coin is worth 1 pfennig.
1 groschen coin is worth 10 pfennigs.
Now, let's set up the equation:
p + 10g = 85
The equation represents the total value in pfennigs. We multiply the number of groschen coins by 10 because each groschen is worth 10 pfennigs. Adding the number of pfennig coins (p) and the number of groschen coins (10g) should give us the total value of 85 pfennigs.
However, since we are looking for a solution where the combined value is 85 pfennigs, we need to consider the restrictions for the number of coins. In this case, we can assume that both p and g are non-negative integers.
Therefore, the equation:
p + 10g = 85
represents the relationship between the number of pfennig coins (p) and groschen coins (g) that have a combined value of 85 pfennigs. So Option B is correct.
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Note the complete question is
From 1990 to 2001, German currency included coins called pfennigs, worth 1 pfennig each, and groschen, worth 10 pfennigs each. Which equation represents the number of pfennig coins, p, and groschen coins, g, that have a combined value of 85 pfennigs?
p + g = 85
p + 10g = 85
10p + g = 85
10(p + g) = 85
Consider the three matrices A = R1 −1 1 0, R2 −1 0 1, R3 6 −2 −3 ; B = C1 2 0 1; C = R1 2 3 1, R2 3 3 1, R3 2 4 1
a) Show by calculating the product AC that C is the inverse matrix of A. Show detailed calculations with an explanation.
b) Hence calculate the solution to the linear system AX = B. Show detailed calculations with an explanation.
The matrices A,B,C,
a) The inverse of Matrix C using the product of matrices is Matrix A.
b) The solution to the linear system AX = B is X = {[5], [7], [5]}.
Matrix A: { [-1 1 0], [-1 0 1], [ 6 -2 -3]}
Matrix B: { [2] [0] [1] }
Matrix C: { C = [ 2 3 1], [ 3 3 1], [ 2 4 1] }
a) To show that C is the inverse matrix of A, let's calculate the product AC and verify if it results in the identity matrix.
Calculating AC:
AC = [-1(2) + 1(3) + 0(2) -1(3) + 1(3) + 0(4) -1(1) + 1(1) + 0(1)]
[-1(2) + 0(3) + 1(2) -1(3) + 0(3) + 1(4) -1(1) + 0(1) + 1(1)]
[6(2) + -2(3) + -3(2) 6(3) + -2(3) + -3(4) 6(1) + -2(1) + -3(1)]
AC = [-2 + 3 + 0 -3 + 3 + 0 -1 + 1 + 0]
[-2 + 0 + 2 -3 + 0 + 4 -1 + 0 + 1]
[12 - 6 - 6 18 - 6 - 12 6 - 2 - 3]
AC = [1 0 0]
[0 1 0]
[0 0 1]
The resulting matrix AC is the identity matrix. Therefore, C is indeed the inverse matrix of A.
b) Now, let's calculate the solution to the linear system AX = B by multiplying both sides by the inverse matrix C:
AX = B
C(AX) = C(B)
Since C is the inverse matrix of A, C(A) is equal to the identity matrix:
I(X) = C(B)
X = C(B)
Now, let's substitute the values of B and C into the equation:
X = [2 3 1] [2]
[3 3 1] [0]
[2 4 1] [1]
X = [(2)(2) + (3)(0) + (1)(1)]
[(3)(2) + (3)(0) + (1)(1)]
[(2)(2) + (4)(0) + (1)(1)]
Therefore, the solution to the linear system AX = B is:
X = [5]
[7]
[5]
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1)Use a First and Second Derivative test to identify the Critical Numbers of f(x)=−2x^3+3x^2+6 then idetitify them as either Max's, Min's or Neither. Make sure to identify where they are by (x,y). Then identify where the point or points of inflection are, (x,y). And finally give the intervals of Concavity for the function. 2) Same directions here as in #1 above for f(x)=sin(x)+cos(x) on [0,4π].
1) The critical numbers are:
Maximum: (1, f(1))
Minimum: (0, f(0))
To identify the critical numbers of the function f(x) = -2x^3 + 3x^2 + 6 and determine whether they are maximums, minimums, or neither,
we need to find the first and second derivatives and analyze their signs.
First, let's find the first derivative:
f'(x) = -6x^2 + 6x
To find the critical numbers, we set the first derivative equal to zero and solve for x:
-6x^2 + 6x = 0
Factor out 6x:
6x(-x + 1) = 0
Set each factor equal to zero:
6x = 0 or -x + 1 = 0
x = 0 or x = 1
So the critical numbers are x = 0 and x = 1.
Next, let's find the second derivative:
f''(x) = -12x + 6
Now we can use the first and second derivative tests.
For x = 0:
f'(0) = -6(0)^2 + 6(0) = 0
f''(0) = -12(0) + 6 = 6
Since the first derivative is zero and the second derivative is positive, we have a local minimum at (0, f(0)).
For x = 1:
f'(1) = -6(1)^2 + 6(1) = 0
f''(1) = -12(1) + 6 = -6
Since the first derivative is zero and the second derivative is negative, we have a local maximum at (1, f(1)).
Therefore, the critical numbers are:
Maximum: (1, f(1))
Minimum: (0, f(0))
2) The function f(x) = sin(x) + cos(x) is concave down in the interval [0, 4π].
Let's find the critical numbers, points of inflection, and intervals of concavity for the function f(x) = sin(x) + cos(x) on the interval [0, 4π].
First, let's find the first derivative:
f'(x) = cos(x) - sin(x)
To find the critical numbers, we set the first derivative equal to zero and solve for x:
cos(x) - sin(x) = 0
Using the trigonometric identity cos(x) = sin(x), we have:
sin(x) - sin(x) = 0
0 = 0
The equation 0 = 0 is always true, so there are no critical numbers in the interval [0, 4π].
Next, let's find the second derivative:
f''(x) = -sin(x) - cos(x)
To find the points of inflection, we set the second derivative equal to zero and solve for x:
-sin(x) - cos(x) = 0
Using the trigonometric identity sin(x) = -cos(x), we have:
-sin(x) + sin(x) = 0
0 = 0
Similarly, the equation 0 = 0 is always true, so there are no points of inflection in the interval [0, 4π].
To determine the intervals of concavity, we need to analyze the sign of the second derivative.
For any value of x in the interval [0, 4π], f''(x) = -sin(x) - cos(x) is negative since both sin(x) and cos(x) are negative in this interval.
Therefore, the function f(x) = sin(x) + cos(x) is concave down in the interval [0, 4π].
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Evaluate integral F .dr where F = <2xy + z, X^2,X> and C
is a path from (1,-1,2) to (2,2,3).
The value of the line integral ∫F · dr along the path C from (1,-1,2) to (2,2,3) is approximately 10.833.'
To evaluate the integral ∫F · dr along the path C from (1,-1,2) to (2,2,3), where F = <2xy + z, x^2, x>, we can parameterize the path C and then perform the line integral using the parameterization.
Let's parameterize the path C by a vector function r(t) = <x(t), y(t), z(t)>, where t ranges from 0 to 1. We need to find the specific parameterization that satisfies the given endpoints (1,-1,2) and (2,2,3).
We can choose the following parameterization:
x(t) = 1 + t
y(t) = -1 + 3t
z(t) = 2 + t
Now, let's find the derivative of r(t) with respect to t:
r'(t) = <1, 3, 1>
The integral ∫F · dr can be written as:
∫[2xy + z, x^2, x] · [dx, dy, dz]
Substituting the parameterization and r'(t) into the integral:
∫[(2(1 + t)(-1 + 3t) + (2 + t)), (1 + t)^2, (1 + t)] · [1, 3, 1] dt
Expanding the dot product and simplifying:
∫[(2 - 2t + 6t^2 + 2 + t), (1 + 2t + t^2), (1 + t)] · [1, 3, 1] dt
Simplifying further:
∫[(9t^2 - t + 4), (t^2 + 2t + 1), (t + 1)] dt
Now, we can integrate each component separately:
∫(9t^2 - t + 4) dt = 3t^3 - (1/2)t^2 + 4t + C1
∫(t^2 + 2t + 1) dt = (1/3)t^3 + t^2 + t + C2
∫(t + 1) dt = (1/2)t^2 + t + C3
Combining the results and adding the constant of integration, we get:
3t^3 - (1/2)t^2 + 4t + C1 + (1/3)t^3 + t^2 + t + C2 + (1/2)t^2 + t + C3
Simplifying and combining the constants of integration:
(3t^3 + (1/3)t^3) - ((1/2)t^2 - t^2) + (4t + t + t) + (C1 + C2 + C3)
The final result of the line integral is:
(10/3)t^3 + (3/2)t^2 + 6t + C
To find the definite integral along the path C from t = 0 to t = 1, we can substitute these values into the expression:
[(10/3)(1)^3 + (3/2)(1)^2 + 6(1)] - [(10/3)(0)^3 + (3/2)(0)^2 + 6(0)]
= (10/3) + (3/2) + 6
= 3.333 + 1.5 + 6
= 10.833
Therefore, the value of the line integral ∫F · dr along the path C from (1,-1,2) to (2,2,3) is approximately 10.833.'
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Determine whether the polygons are always, sometimes, or never similar. Explain your reasoning.
a trapezoid and a parallelogram
A trapezoid and parallelogram can be sometimes similar, as they can have the same shape but different sizes.
1. Similar polygons have the same shape but can be different sizes.
2. A trapezoid and a parallelogram can have the same shape, but their angles and side lengths may differ.
3. Therefore, they can be sometimes similar, depending on their specific measurements.
A trapezoid and parallelogram can be sometimes similar, as they can have the same shape but different sizes. Polygons have the same shape but can be different sizes.
A trapezoid and a parallelogram can have the same shape, but their angles and side lengths may differ.Therefore, they can be sometimes similar, depending on their specific measurements.
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When given two points to determine the equation of a line, either of the given points can be used to put the equation into point-slope form.
To put the equation of a line into point-slope form, use either of the given points and the slope: y - y1 = m(x - x1).
When given two points to determine the equation of a line, point-slope form can be used. Point-slope form is represented as y - y1 = m(x - x1), where (x1, y1) denotes one of the given points, and m represents the slope of the line. To convert the equation into point-slope form, you can select either of the points and substitute its coordinates into the equation along with the calculated slope.
This form allows you to easily express a linear relationship between variables and graph the line accurately. It is a useful tool in various applications, such as analyzing data, solving problems involving lines, or determining the equation of a line given two known points.
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Chris Lynch plans to invest $500 into a money market account. Find the interest rate that is needed for the money to grow to $1,800 in 14 years if the interest is compounded quarterly.
Let's first find the number of quarterly periods in 14 years:14 years × 4 quarters per year = 56 quarters Next, let's use the formula A = P(1 + r/n)nt where: A = final amount P = principal r = annual interest rate (as a decimal)n = number of times compounded per year t = time in years.
Therefore, the formula becomes:$1,800 = $500(1 + r/4)^(4×14/1)$1,800/$500 = (1 + r/4)^56$3.6 = (1 + r/4)^56Now take the 56th root of both sides:56th root of 3.6 ≈ 1 + r/4r/4 ≈ 0.0847r ≈ 0.0847 × 4r ≈ 0.3388
Therefore, the interest rate that is needed for the money to grow to $1,800 in 14 years if the interest is compounded quarterly is approximately 33.88%.
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An equation of a line through (−1,2) which is perpendicular to the line y=4x+1 has slope: and y intercept at:
The slope of the line perpendicular to y = 4x + 1 is -1/4, and the y-intercept is 9/4.
To find the equation of a line perpendicular to y = 4x + 1, we need to determine the negative reciprocal of the slope of the given line.
1. Given line: y = 4x + 1
2. The slope of the given line is 4. The negative reciprocal of 4 is -1/4.
3. So, the slope of the perpendicular line is -1/4.
4. We also know that the line passes through the point (-1, 2). We can use this point to find the y-intercept of the perpendicular line.
5. The equation of a line can be written in the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
6. Substitute the slope (-1/4), the coordinates (-1, 2), and solve for b:
2 = (-1/4)(-1) + b
2 = 1/4 + b
b = 2 - 1/4
b = 8/4 - 1/4
b = 7/4
7. Therefore, the equation of the line perpendicular to y = 4x + 1 is y = (-1/4)x + 7/4.
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Consider the curve described by the vector function r(t)=t cos ti+t sin tj+2tk. (a) Show that motion along this curve occurs at an increasing speed as t>0 increases. Hint: The speed at a point is given by the length of the tangent vector at that point.) (b) Find parametric equations for the line tangent to this curve at the point (0,π/2,π).
(a) To show that motion along the curve described by the vector function [tex]\( r(t) = t \cos(t)i + t \sin(t)j + 2tk \)[/tex] occurs at an increasing speed as t > 0 increases, we need to find the speed function
(a) The speed at a point on the curve is given by the magnitude of the tangent vector at that point. The derivative of the position vector r(t) with respect to t gives the tangent vector r'(t). The speed function is given by r'(t) , the magnitude of r'(t). By finding the derivative of the speed function with respect to t and showing that it is positive for t > 0 , we can conclude that motion along the curve occurs at an increasing speed as t increases.
(b) To find the parametric equations for the line tangent to the curve at the point [tex]\((0, \frac{\pi}{2}, \pi)\)[/tex], we need to find the derivative of the vector function \( r(t) \) and evaluate it at that point.
The derivative is given by[tex]\( r'(t) = \frac{d}{dt} (t \cos(t)i + t \sin(t)j + 2tk) \)[/tex]. Evaluating r'(t) at t = 0, we obtain the direction vector of the tangent line. Using the point-direction form of the line equation, we can write the parametric equations for the line tangent to the curve at the given point.
In summary, to show that motion along the curve occurs at an increasing speed as t > 0 increases, we analyze the speed function. To find the parametric equations for the line tangent to the curve at the point[tex]\((0, \frac{\pi}{2}, \pi)\)[/tex], we differentiate the vector function and evaluate it at that point.
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Partial fraction division: \[ \frac{x+2}{x^{4}-3 x^{3}+x^{2}+3 x-2} \]
To perform partial fraction decomposition on the given rational function, we start by factoring the denominator. The denominator
x
4
−
3
x
3
+
x
2
+
3
x
−
2
x
4
−3x
3
+x
2
+3x−2 can be factored as follows:
x
4
−
3
x
3
+
x
2
+
3
x
−
2
=
(
x
2
−
2
x
+
1
)
(
x
2
+
x
−
2
)
x
4
−3x
3
+x
2
+3x−2=(x
2
−2x+1)(x
2
+x−2)
Now, we can express the rational function as a sum of partial fractions:
x
+
2
x
4
−
3
x
3
+
x
2
+
3
x
−
2
=
A
x
2
−
2
x
+
1
+
B
x
2
+
x
−
2
x
4
−3x
3
+x
2
+3x−2
x+2
=
x
2
−2x+1
A
+
x
2
+x−2
B
To find the values of
A
A and
B
B, we need to find a common denominator for the fractions on the right-hand side. Since the denominators are already irreducible, the common denominator is simply the product of the two denominators:
x
4
−
3
x
3
+
x
2
+
3
x
−
2
=
(
x
2
−
2
x
+
1
)
(
x
2
+
x
−
2
)
x
4
−3x
3
+x
2
+3x−2=(x
2
−2x+1)(x
2
+x−2)
Now, we can equate the numerators on both sides:
x
+
2
=
A
(
x
2
+
x
−
2
)
+
B
(
x
2
−
2
x
+
1
)
x+2=A(x
2
+x−2)+B(x
2
−2x+1)
Expanding the right-hand side:
x
+
2
=
(
A
+
B
)
x
2
+
(
A
+
B
)
x
+
(
−
2
A
+
B
)
x+2=(A+B)x
2
+(A+B)x+(−2A+B)
By comparing coefficients on both sides, we obtain the following system of equations:
A
+
B
=
1
A+B=1
A
+
B
=
1
A+B=1
−
2
A
+
B
=
2
−2A+B=2
Solving this system of equations, we find that
A
=
1
3
A=
3
1
and
B
=
2
3
B=
3
2
.
Therefore, the partial fraction decomposition of the given rational function is:
x
+
2
x
4
−
3
x
3
+
x
2
+
3
x
−
2
=
1
3
(
x
2
−
2
x
+
1
)
+
2
3
(
x
2
+
x
−
2
)
x
4
−3x
3
+x
2
+3x−2
x+2
=
3(x
2
−2x+1)
1
+
3(x
2
+x−2)
2
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c. Using systematic random sampling, every seventh dealer is selected starting with the fourth dealer in the list. Which dealers are included in the sample
The fourth, eleventh, eighteenth, twenty-fifth, and so on, dealers in the list would be included in the sample.
Using systematic random sampling, every seventh dealer is selected starting with the fourth dealer in the list. The process continues until the desired sample size is reached or until all dealers have been included in the sample.
Since the question does not specify the total number of dealers in the list or the desired sample size, it is not possible to provide specific dealer numbers that are included in the sample.
However, based on the given sampling method, the sample would consist of dealers at regular intervals of seven starting from the fourth dealer in the list.
This means that the fourth, eleventh, eighteenth, twenty-fifth, and so on, dealers in the list would be included in the sample.
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where do the graphs of the linear equations 10x 12y = 14 and 5x 6y = 7 intersect?
The graphs of the linear equations 10x - 12y = 14 and 5x - 6y = 7 intersect along the entire line represented by the equations.
To find the point of intersection between the graphs of the linear equations 10x - 12y = 14 and 5x - 6y = 7, we can solve the system of equations simultaneously.
First, let's solve the second equation for x:
5x - 6y = 7
5x = 6y + 7
x = (6y + 7) / 5
Next, substitute this expression for x into the first equation:
10x - 12y = 14
10((6y + 7) / 5) - 12y = 14
12y + 14 - 12y = 14
14 = 14
The equation 14 = 14 is always true. This indicates that the two equations represent the same line and are coincident. Therefore, the graphs of the two equations overlap and intersect at all points along the line defined by the equations.
In summary, the graphs of the linear equations 10x - 12y = 14 and 5x - 6y = 7 intersect along the entire line represented by the equations.
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What is correct form of the particular solution associated with the differential equation y ′′′=8? (A) Ax 3 (B) A+Bx+Cx 2 +Dx 3 (C) Ax+Bx 2 +Cx 3 (D) A There is no correct answer from the given choices.
To find the particular solution associated with the differential equation y′′′ = 8, we integrate the equation three times.
Integrating the given equation once, we get:
y′′ = ∫ 8 dx
y′′ = 8x + C₁
Integrating again:
y′ = ∫ (8x + C₁) dx
y′ = 4x² + C₁x + C₂
Finally, integrating one more time:
y = ∫ (4x² + C₁x + C₂) dx
y = (4/3)x³ + (C₁/2)x² + C₂x + C₃
Comparing this result with the given choices, we see that the correct answer is (B) A + Bx + Cx² + Dx³, as it matches the form obtained through integration.
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1. (10 points) Find \( \int x \arctan x d x \)
To find \( \int x \arctan x \, dx \), we can use integration by parts. By choosing \( u = \arctan x \) and \( dv = x \, dx \), we can differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \).
Applying the integration by parts formula, we can then evaluate the integral.
To evaluate \( \int x \arctan x \, dx \), we choose \( u = \arctan x \) and \( dv = x \, dx \). Taking the derivative of \( u \) gives \( du = \frac{1}{1 + x^2} \, dx \), and integrating \( dv \) yields \( v = \frac{1}{2}x^2 \). Applying the integration by parts formula:
\( \int u \, dv = uv - \int v \, du \)
we have:
\( \int x \arctan x \, dx = \frac{1}{2}x^2 \arctan x - \int \frac{1}{2}x^2 \cdot \frac{1}{1 + x^2} \, dx \)
Simplifying the integral, we get:
\( \int x \arctan x \, dx = \frac{1}{2}x^2 \arctan x - \frac{1}{2} \int \frac{x^2}{1 + x^2} \, dx \)
The remaining integral on the right-hand side can be evaluated using a substitution or other integration techniques.
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determine the interval in which solutions are sure to exist. y′′′ ty'' t^2y'=ln(t)
Therefore, the interval in which solutions are sure to exist is (0, ∞).
To determine the interval in which solutions are sure to exist for the given differential equation, we need to consider any restrictions or limitations imposed by the equation itself.
In this case, the given differential equation is:
y′′′ ty'' t^2y'=ln(t)
The equation involves logarithm function ln(t), which is not defined for t ≤ 0. Therefore, the interval in which solutions are sure to exist is t > 0.
In other words, solutions to the given differential equation can be found for values of t that are strictly greater than 0.
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Let A and B be n×n matrices with det(A)=6 and det(B)=−1. Find det(A 7
B 3
(B T
A 8
) −1
A T
) Answer:
Given,A and B be n×n matrices with det(A)=6 and det(B)=−1. Find det(A7B3(BTA8)−1AT)So, we have to find the value of determinant of the given expression.A7B3(BTA8)−1ATAs we know that:(AB)T=BTATWe can use this property to find the value of determinant of the given expression.A7B3(BTA8)−1AT= (A7B3) (BTAT)−1( AT)Now, we can rearrange the above expression as: (A7B3) (A8 BT)−1(AT)∴ (A7B3) (A8 BT)−1(AT) = (A7 A8)(B3BT)−1(AT)
Let’s first find the value of (A7 A8):det(A7 A8) = det(A7)det(A8) = (det A)7(det A)8 = (6)7(6)8 = 68 × 63 = 66So, we got the value of (A7 A8) is 66.
Let’s find the value of (B3BT):det(B3 BT) = det(B3)det(BT) = (det B)3(det B)T = (−1)3(−1) = −1So, we got the value of (B3 BT) is −1.
Now, we can substitute the values of (A7 A8) and (B3 BT) in the expression as:(A7B3(BTA8)−1AT) = (66)(−1)(AT) = −66det(AT)Now, we know that, for a matrix A, det(A) = det(AT)So, det(AT) = det(A)∴ det(A7B3(BTA8)−1AT) = −66 det(A)We know that det(A) = 6, thus∴ det(A7B3(BTA8)−1AT) = −66 × 6 = −396.Hence, the determinant of A7B3(BTA8)−1AT is −396. Answer more than 100 words:In linear algebra, the determinant of a square matrix is a scalar that can be calculated from the elements of the matrix.
If we have two matrices A and B of the same size, then we can define a new matrix as (AB)T=BTA. With this property, we can find the value of the determinant of the given expression A7B3(BTA8)−1AT by rearranging the expression. After the rearrangement, we need to find the value of (A7 A8) and (B3 BT) to substitute them in the expression.
By using the property of determinant that the determinant of a product of matrices is equal to the product of their determinants, we can calculate det(A7 A8) and det(B3 BT) easily. By putting these values in the expression, we get the determinant of A7B3(BTA8)−1AT which is −396. Hence, the solution to the given problem is concluded.
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The profit for a product is given by
P(x)=11x−5063,
where x is the number of units produced and sold. Find the
marginal profit for the product.
The marginal profit for the product is $11 per unit. This indicates the rate at which the profit changes with respect to the number of units produced and sold.
The profit function is given by P(x) = 11x - 5063, where x represents the number of units produced and sold. To find the marginal profit, we need to find the derivative of P(x) with respect to x.
Taking the derivative of P(x), we have dP/dx = d/dx (11x - 5063).
Differentiating each term separately, we get dP/dx = 11.
The derivative of the profit function is a constant value of 11, which represents the marginal profit. This means that for every additional unit produced and sold, the profit increases by $11.
Therefore, the marginal profit for the product is $11 per unit. This indicates the rate at which the profit changes with respect to the number of units produced and sold.
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a sheet of gold weighing 10.0 g and at a temperature of 18.0°c is placed fl at on a sheet of iron weighing 20.0 g and at a temperature of 55.6°c. what is the fi nal temperature of the combined metals?
When different metals with different temperatures are placed together, they tend to exchange heat until the temperature becomes equal. This phenomenon is known as Thermal Equilibrium.
The final temperature of the combined metals can be calculated using the following formula:
Q = m * c * ∆T
Where,Q = Heat exchanged by metals m = Mass of metals c = Specific Heat of metal∆T = Change in temperature
Assuming no heat is lost to the surroundings, we can say that the Heat lost by the hot iron is equal to the Heat gained by the cold gold.
Hence, m1 * c1 * ∆T1 = m2 * c2 * ∆T2.
Rearranging the equation,
we get ∆T = (m1 * c1 * ∆T1) / (m2 * c2).
Now substituting the values, we get;For gold, m = 10 g, c = 0.129 J/g°C, ∆T = (Tfinal - 18°C).
For iron, m = 20 g, c = 0.449 J/g°C, ∆T = (55.6 - Tfinal).
We get ∆T = (10 * 0.129 * (Tfinal - 18)) / (20 * 0.449) = (1.29 * (Tfinal - 18)) / 8.98.
Now equating the two, we get (Tfinal - 18) / 8.98 = (55.6 - Tfinal) / 20.
Solving the equation,
we get Tfinal = (55.6 * 8.98 + 18 * 20) / (8.98 + 20) = 30.18°C.
Hence the final temperature of the combined metals is 30.18°C.
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Sketch the graph of the function. y=ln(x+5)
Given: y = ln(x + 5)To sketch the graph of the function, y = ln(x + 5) the following steps need to be followed:Step 1: Finding the domain of the functionFor any natural logarithmic function, the argument must be greater than zero: x + 5 > 0x > -5.
The domain of the function is (-5, ∞)Step 2: Finding the intercepts of the functionTo find the y-intercept, let x = 0y = ln(0 + 5) = ln(5)To find the x-intercept, let y = 0.0 = ln(x + 5)x + 5 = e0 = 1x = -5The intercepts are (0, ln5) and (-5, 0)Step 3: Finding the asymptotes To find the vertical asymptote, solve for x in the equation: x + 5 = 0x = -5 The vertical asymptote is x = -5.
The horizontal asymptote can be found by taking the limit as x approaches infinity:limx → ∞ ln(x + 5) = ∞The horizontal asymptote is y = ∞Step 4: Sketch the graphUsing the above information, sketch the graph of the function:The graph is shown below:Answer: The graph of the function y = ln(x + 5) is shown below:
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Consider the differential equation
y′′ + 3y′ − 10y = 0.
(a) Find the general solution to this differential equation.
(b) Now solve the initial value problem corresponding to y(0) = 2 and y′(0) = 10
The answer of the given question based on the differential equation is , the solution of the given initial value problem is: y = (-16/7)e-5t + (30/7)e2t
The given differential equation is:
y'' + 3y' - 10y = 0
(a) Find the general solution to this differential equation.
The auxiliary equation is:
r2 + 3r - 10 = 0
Factorizing the above equation, we get:
(r + 5)(r - 2) = 0r = -5 or r = 2
Thus, the general solution of the given differential equation is given by:
y = c1e-5t + c2e2t
(b) Solve the initial value problem corresponding to y(0) = 2 and y′(0) = 10
To solve the initial value problem, we need to find the values of c1 and c2.
Substituting t = 0 and y = 2 in the above general solution, we get:
2 = c1 + c2 ........(1)
Differentiating the above general solution, we get:
y' = -5c1e-5t + 2c2e2t
Substituting t = 0 and y' = 10 in the above equation, we get:
10 = -5c1 + 2c2 .........(2)
On solving equations (1) and (2), we get:
c1 = -16/7 and c2 = 30/7
Thus, the solution of the given initial value problem is: y = (-16/7)e-5t + (30/7)e2t
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