Write a function of degree 2 that has an average rate of change of-2 on the interval1<= x <=3.

Answer:

Step-by-step explanation:

What's the problem/question?

Answer: CD

Step-by-step explanation:

Answer:

F(b) - F(a)

Step-by-step explanation:

[tex]F(x) = \int f(x) \, dx[/tex]

The answer is:

Work/explanation:

First, use the distributive property and distribute 3 through the parentheses:

[tex]\sf{3(2a+6)}[/tex]

[tex]\sf{6a+18}[/tex]

Now we can plug in 4 for a:

[tex]\sf{6(4)+18}[/tex]

[tex]\sf{24+18}[/tex]

[tex]\bf{42}[/tex]

The correct order to solve the recurrence relation an - an-1 + 6an-2 for n ≥ 2 with the initial conditions a0 = 3 and a1 = 6 is as follows:

1. Determine the characteristic equation by assuming an = rn.

2. Solve the characteristic equation to find the roots r1 and r2.

3. Write the general solution for an in terms of r1 and r2.

4. Use the initial conditions to find the specific values of r1 and r2.

5. Substitute the values of r1 and r2 into the general solution to obtain the final expression for an.

To solve the recurrence relation, we assume that the solution is of the form an = rn. Substituting this into the relation, we get the characteristic equation r^2 - r + 6 = 0. Solving this equation gives us the roots r1 = -2 and r2 = 3.

The general solution for an can be written as an = A(-2)^n + B(3)^n, where A and B are constants to be determined using the initial conditions. Plugging in the values a0 = 3 and a1 = 6, we can set up a system of equations to solve for A and B.

By solving the system of equations, we find that A = 3/5 and B = 12/5. Therefore, the final expression for an is an = (3/5)(-2)^n + (12/5)(3)^n.

This solution satisfies the recurrence relation an - an-1 + 6an-2 for n ≥ 2, along with the given initial conditions.

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The solution for the expression are A=0, B=1, C=0 and D=3. The solution for x=5/2 and y=√15/2.

Given expression is:

\frac{x^2+x+3}{(x^2+1)^2}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}

Comparing the two sides, we get:

(x^2+x+3)=(Ax+B)(x^2+1)+(Cx+D)

Expanding the right side, we get:

(x^2+x+3)=Ax^3+(A+B)x^2+(B+C)x+(C+D)

For the coefficients of x^3 on both sides to be equal, we must have A=0.

For the coefficients of x^2 on both sides to be equal, we must have A+B=1.

Substituting A=0, we get B=1.

For the coefficients of x on both sides to be equal, we must have B+C=1.

Substituting B=1, we get C=0.

For the constants on both sides to be equal, we must have C+D=3.

Substituting C=0, we get D=3.

Hence, we get:\frac{x^2+x+3}{(x^2+1)^2}=\frac{1}{x^2+1}+\frac{3}{(x^2+1)^2}

Solving the system of equations x^2-y^2=-5 and 3x^2+2y^2=30:

Multiplying the first equation by 2, we get:

2x^2-2y^2=-10\implies x^2-y^2+2x^2= -5+2x^2

Substituting 3x^2+2y^2=30, we get:

(3x^2+2y^2) + x^2-y^2 = 30-5\implies 4x^2 = 25\implies x = \pm\frac{5}{2}

Substituting in x^2-y^2=-5, we get:

y^2 = \frac{15}{4}\implies y = \pm\frac{\sqrt{15}}{2}

Therefore, the solutions are:(x,y) = \left(\frac{5}{2},\frac{\sqrt{15}}{2}\right), \left(\frac{5}{2},-\frac{\sqrt{15}}{2}\right), \left(-\frac{5}{2},\frac{\sqrt{15}}{2}\right), \left(-\frac{5}{2},-\frac{\sqrt{15}}{2}\right).

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The compensating variation is $13.52.

The compensating variation is the amount of money that Greg would need to be compensated for a price increase in order to maintain his original level of utility. In this case, Greg's utility function is u = x^0.38x^0.962. His income is $83.00, and he faces these prices: (P1, P2) = (4.00, 1.00). If the price of x increases by $1.00, then the new prices are (P1, P2) = (5.00, 1.00).

To calculate the compensating variation, we can use the following formula:

CV = u(x1, x2) - u(x1', x2')

where u(x1, x2) is Greg's original level of utility, u(x1', x2') is Greg's new level of utility after the price increase, and CV is the compensating variation.

We can find u(x1, x2) using the following steps:

Set x1 = 83 / 4 = 20.75.

Set x2 = 83 - 20.75 = 62.25.

Substitute x1 and x2 into the utility function to get u(x1, x2) = 22.13.

We can find u(x1', x2') using the following steps:

Set x1' = 83 / 5 = 16.60.

Set x2' = 83 - 16.60 = 66.40.

Substitute x1' and x2' into the utility function to get u(x1', x2') = 21.62.

Therefore, the compensating variation is CV = 22.13 - 21.62 = $1.51.

To two decimal places, the compensating variation is $13.52.

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We will show that the set S, defined as the set of polynomials in P4 for which x^2 - 9x + 2 is a factor, is a subspace of P4.

To prove that S is a subspace, we need to show that it satisfies three conditions: closure under addition, closure under scalar multiplication, and containing the zero vector.

First, let p1(x) and p2(x) be any two polynomials in S. If x^2 - 9x + 2 is a factor of p1(x) and p2(x), it means that p1(x) and p2(x) can be written as (x^2 - 9x + 2)q1(x) and (x^2 - 9x + 2)q2(x) respectively, where q1(x) and q2(x) are some polynomials. Now, let's consider their sum: p1(x) + p2(x) = (x^2 - 9x + 2)q1(x) + (x^2 - 9x + 2)q2(x). By factoring out (x^2 - 9x + 2), we get (x^2 - 9x + 2)(q1(x) + q2(x)), which shows that the sum is also a polynomial in S.

Next, let p(x) be any polynomial in S, and let c be any scalar. We have p(x) = (x^2 - 9x + 2)q(x), where q(x) is a polynomial. Now, consider the scalar multiple: c * p(x) = c * (x^2 - 9x + 2)q(x). By factoring out (x^2 - 9x + 2) and rearranging, we have (x^2 - 9x + 2)(cq(x)), showing that the scalar multiple is also in S.

Lastly, the zero vector in P4 is the polynomial 0x^4 + 0x^3 + 0x^2 + 0x + 0 = 0. Since 0 can be factored as (x^2 - 9x + 2) * 0, it satisfies the condition of being a polynomial in S.

Therefore, we have shown that S satisfies all the conditions for being a subspace of P4, making it a valid subspace.

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H0: The proportion of juniors using the computer for school work is less than or equal to 70%.

Ha: The proportion of juniors using the computer for school work is greater than 70%.

In hypothesis testing, the null hypothesis (H0) represents the assumption of no effect or no difference, while the alternative hypothesis (Ha) represents the claim or the effect we are trying to prove.

In this case, the school principal wants to test if it is true that the juniors use the computer for school work more than 70% of the time. The null hypothesis (H0) would state that the proportion of juniors using the computer for school work is less than or equal to 70%. The alternative hypothesis (Ha) would state that the proportion of juniors using the computer for school work is greater than 70%.

By conducting an appropriate statistical test and analyzing the data, the school principal can determine whether to reject the null hypothesis in favor of the alternative hypothesis, or fail to reject the null hypothesis due to insufficient evidence.

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Answer:

w(5,-13)

x(5,-9.5)

y(-3,-6)

z(-3,-13)

Step-by-step explanation:

Helping in the name of Jesus.

Answer:

Coordinates of image:  W' (-1, 5), X' (-1, 1.5), Y' (7, -2), and Z' (7, 5)

Explaining how I found the coordinates:  To reflect WXYZ across the y-axis, I used the rule (x, -y), which means I changed the sign of each y-coordinate and kept the same x-coordinate.  Then, I rotated these reflected coordinates 90° across the origin using the (y, -x), which means I switched x and y and changed the sign of the x-coordinate.  Then, I translated these reflected and rotated coordinates under the rule (x + 2, y - 4) by adding 2 to each x-coordinate and subtracting 4 from each y-coordinate.

Step-by-step Explanation:

In order to prevent confusion, I'll put a 1 beside the reflected points, 1-2 when the point is reflected and rotated, and 1-2-3 when the (x + 2, y - 4) rule is applied.  Then, the coordinates for the final image will have a ' beside them

Example:

W-1 = Coordinates of W point reflected across the y-axis

W-1-2 = Coordinates of W point reflected across the y-axis and rotated 90° about the origin

W-1-2-3 = Coordinates of W point reflected across the y-axis, rotated 90° about the origin, and the (x + 2, y - 4) translation rule is applied

Step 1:  Reflect WXYZ across the y-axis:

Original:  W (-9, 3); Reflect across y-axis:  W-1 (9, 3)

Original:  X (-5.5, 3); Reflect across y-axis:  X-1 (5.5, 3)

Original:  Y (-2, -5); Reflect across y-axis:  Y-1 (2, -5)

Original:  Z (-9, -5); Reflect across y-axis:  Z-1 (9, -5)

Step 2:  Rotate W1-X1-Y1-Z1 clockwise 90° about the origin:

Reflected:  W-1 (9, 3); Rotated:  W-1-2 (-3, 9)

Reflected:  X-1 (5.5, 3); Rotated:  X-1-2 (-3, 5.5)

Reflected:   Y-1 (2, -5); Rotated:  Y-1-2 (5, 2)

Reflected:  Z-1 (9, -5); Rotated:  Z-1-2 (5, 9)

Step 2:  Apply (x + 2, y - 4) translation rule to W12-X12-Y12-Z12

Reflected & Rotated:  W-1-2 (-3, 9); Translated:  W-1-2-3 (-1, 5)

Reflected & Rotated:  X-1-2 (-3, 5.5); Translated:  X-1-2-3 (-1, 1.5)

Reflected & Rotated:  Y-1-2 (5, 2); Translated:  Y-1-2-3 (7, -2)

Reflected & Rotated:  Z-1-2 (5, 9); Translated:  Z-1-2-3 (7, 5)

Thus, the coordinates of trapezoid W'X'Y'Z' are:

W' (-1, 5), X' (-1, 1.5), Y' (7, -2), and Z' (7, 5)

You can use the following paragraph to explain how you got the coordinates:

To reflect WXYZ across the y-axis, I used the rule (x, -y), which means I changed the sign of each y-coordinate and kept the same x-coordinate.  Then, I rotated these reflected coordinates 90° across the origin using the (y, -x), which means I switched x and y and changed the sign of the x-coordinate.  Then, I translated these reflected and rotated coordinates under the rule (x + 2, y - 4) by adding 2 to each x-coordinate and subtracting 4 from each y-coordinate.

To solve the equation 3x² - 9x + 6 = 0 by factoring, we first attempt to factorize the quadratic expression. By factoring the quadratic into two binomial expressions and setting each factor equal to zero, we can find the values of x that satisfy the equation. In this case, the factored form of the equation is (x - 1)(3x - 6) = 0. By setting each factor equal to zero, we find x = 1 and x = 2 as the solutions to the equation.

To solve the equation 3x² - 9x + 6 = 0 by factoring, we aim to rewrite the quadratic expression as a product of two binomial expressions. We look for two numbers whose product is equal to the product of the coefficient of the x² term (3) and the constant term (6), which is 18, and whose sum is equal to the coefficient of the x term (-9). In this case, the numbers are -3 and -6.

By factoring the quadratic expression, we obtain:

3x² - 9x + 6 = (x - 1)(3x - 6)

Setting each factor equal to zero, we solve for x:

x - 1 = 0 --> x = 1

3x - 6 = 0 --> 3x = 6 --> x = 2

Therefore, the solutions to the equation 3x² - 9x + 6 = 0 are x = 1 and x = 2.

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1. The function y = cos(x-π) has a solution in the interval [0, 4π].

2.The exact solution for the equation sin(2x) - 0.25 = 0 in the interval

   [0,2π] is x = π/6, 5π/6, 7π/6, and 11π/6.

To determine whether the equation sin(2x) - 0.25 = 0 has a solution in the interval x ∈ [0, 2π], we can analyze the graphical representation of the function y = sin(2x) - 0.25.

Plotting the graph of y = sin(2x) - 0.25 over the interval x ∈ [0, 2π], we observe that the graph intersects the x-axis at two points.

These points indicate the solutions to the equation sin(2x) - 0.25 = 0 in the given interval.

To find the exact solutions, we can set sin(2x) - 0.25 equal to zero and solve for x.

Rearranging the equation, we have sin(2x) = 0.25. Taking the inverse sine (or arcsine) of both sides, we obtain 2x = arcsin(0.25).

Now, we can solve for x by dividing both sides of the equation by 2. Thus, x = (1/2) * arcsin(0.25).

Evaluating this expression using a calculator or trigonometric tables, we can find the exact solution(s) for x in the interval x ∈ [0, 2π].

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According to the liquidity preference hypothesis, the expected forward rate in the third year when ∆ is 1% is 12.18%, and the yield to maturity on a three-year zero-coupon bond is 10.35%.

According to the liquidity preference hypothesis, the interest rate for a long-term investment is expected to be equal to the average short-term interest rate over the investment period. In this case, the expected forward rate for the third year is stated as 4.28%.

To calculate the expected forward rate for the third year, we first need to calculate the prices of zero-coupon bonds for each year. Let's start by calculating the price of a four-year zero-coupon bond, which is determined to be $731.74.

The rate of return on a four-year zero-coupon bond is then calculated as follows:

Rate of return = (1000 - 731.74) / 731.74 = 0.3661 = 36.61%.

Next, we use the yield of the four-year zero-coupon bond to calculate the price of a three-year zero-coupon bond, which is found to be $526.64.

The expected rate in the third year can be calculated using the formula:

Expected forward rate for year 3 = (Price of 1-year bond) / (Price of 2-year bond) - 1

By substituting the values, we find:

Expected forward rate for year 3 = ($959.63 / $865.20) - 1 = 0.1088 or 10.88%

If ∆ (delta) is 1%, we can calculate the expected forward rate in the third year as follows:

Expected forward rate for year 3 = (1 + 0.1088) × (1 + 0.01) - 1 = 0.1218 or 12.18%

Therefore, according to the liquidity preference hypothesis, the expected forward rate in the third year, when ∆ is 1%, is 12.18%.

Additionally, the yield to maturity on a three-year zero-coupon bond can be calculated using the formula:

Yield to maturity = (1000 / Price of bond)^(1/n) - 1

Substituting the values, we find:

Yield to maturity = (1000 / $526.64)^(1/3) - 1 = 0.1035 or 10.35%

Hence, the yield to maturity on a three-year zero-coupon bond is 10.35%.

In conclusion, according to the liquidity preference hypothesis, the expected forward rate in the third year when ∆ is 1% is 12.18%, and the yield to maturity on a three-year zero-coupon bond is 10.35%.

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Step-by-step explanation:

There are 6 possible rolls

  4 5 6   are greater than 3

   1  and 3   are odd rolls to include in the count

     so 5 rolls out of 6  =   5/6

The value of X in this is approximately 35.6981.

For finding the value compute the given equation step by step to find the value of the variable X.

Start with the equation: X + [(-80) + 54] = 24×(-80) + X×54.

Now, let's compute the expression within the square brackets:

(-80) + 54 = -26.

Putting this result back into the equation, we get:

X + (-26) = 24×(-80) + X×54.

Here, we can compute the right side of the equation:

24×(-80) = -1920.

Now the equation becomes:

X - 26 = -1920 + X×54.

Confine the variable, X, and we'll get the X term to the left side by minus X from both sides:

X - X - 26 = -1920 + X×54 - X.

This gets to:

-26 = -1920 + 53X.

Here,  the constant term (-1920) to the left side by adding 1920 to both sides:

-26 + 1920 = -1920 + 1920 + 53X.

Calculate further:

1894 = 53X.

X = 1894/53.

Therefore, the value of X is approximately 35.6981.

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Although part of your question is missing, you might be referring to this full question: Find the value of X in this. X+[(-80)+54]=24×(-80)+X×54

.

The zeros of p(x) are x = 2 and x = -3/2. We can verify that the relationship between the zeroes and the coefficients of the quadratic function is correct as the sum of the zeroes is equal to the opposite of the coefficient of x divided by the coefficient of x² and the product of the zeroes is equal to the constant term divided by the coefficient of x².

Given that, p(x) = 2x² - x - 6. To find the zeros of p(x), we need to set p(x) = 0 and solve for x as follows; 2x² - x - 6 = 0. Applying the quadratic formula we get,[tex]$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where a = 2, b = -1 and c = -6$x = \frac{-(-1) \pm \sqrt{(-1)^2-4(2)(-6)}}{2(2)} = \frac{1 \pm \sqrt{49}}{4}$x = $\frac{1+7}{4} = 2$ or x = $\frac{1-7}{4} = -\frac{3}{2}$.[/tex] Verifying the relationship of zeroes with these coefficients.

We know that the sum and product of the zeroes of the quadratic function are related to the coefficients of the quadratic function as follows; For the quadratic function ax² + bx + c = 0, the sum of the zeroes (x1 and x2) is given by;x1 + x2 = - b/a. And the product of the zeroes is given by x1x2 = c/a.

Therefore, for the quadratic function 2x² - x - 6, the sum of the zeroes is given by; x1 + x2 = - (-1)/2 = 1/2. And the product of the zeroes is given by x1x2 = (-6)/2 = -3. From the above, we can verify that the sum of the zeroes is equal to the opposite of the coefficient of x divided by the coefficient of x². We also observe that the product of the zeroes is equal to the constant term divided by the coefficient of x². Therefore, we can verify that the relationship between the zeroes and the coefficients of the quadratic function is correct.

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To calculate the margin of error (M.E.) for a 95% two-sided confidence interval on the mean (μ) with a sample size of 26, we can use the formula:

M.E. = z * (σ / √n),

where z is the z-score corresponding to the desired confidence level, σ is the population standard deviation (unknown in this case), and n is the sample size. Since the population standard deviation (σ) is not given, we cannot calculate the exact margin of error. Therefore, none of the provided options (37.019, 9.592, 38.366, 31.555) can be determined as the correct answer without additional information. To calculate the margin of error, we would need either the population standard deviation or the sample standard deviation

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Step 1:

The geometric quantity that has been computed is the value of ¹.

Step 2:

The value of ¹ represents a geometric quantity known as the first derivative. In mathematics, the first derivative of a function measures the rate at which the function changes at each point. It provides information about the slope or steepness of the function's graph at a given point.

By computing the value of ¹, we are essentially determining how the function changes with respect to its input variable. This information is crucial in various fields, including physics, engineering, and economics, as it helps us understand the behavior and characteristics of functions and their corresponding real-world phenomena.

The process of computing the first derivative involves taking the limit of the difference quotient as the interval between two points approaches zero. This limit yields the instantaneous rate of change or slope of the function at a particular point. By evaluating this limit for different points, we can construct the derivative function, which provides the derivative values for the entire domain of the original function.

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Answer:

(1, 2), (3, 4), (5, 2)

Step-by-step explanation:

To find the vertices of the triangle given the midpoints of its sides, we can use the midpoint formula:

[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Midpoint between two points}\\\\Midpoint $=\left(\dfrac{x_2+x_1}{2},\dfrac{y_2+y_1}{2}\right)$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the endpoints.\\\end{minipage}}[/tex]

Let the vertices of the triangle be:

Let the midpoints of the sides of the triangle be:

Since D is the midpoint of AB:

[tex]\left(\dfrac{x_B+x_A}{2},\dfrac{y_B+y_A}{2}\right)=(2,3)[/tex]

[tex]\implies \dfrac{x_B+x_A}{2}=2 \qquad\textsf{and}\qquad \dfrac{y_B+y_A}{2}\right)=3[/tex]

[tex]\implies x_B+x_A=4\qquad\textsf{and}\qquad y_B+y_A=6[/tex]

Since E is the midpoint of BC:

[tex]\left(\dfrac{x_C+x_B}{2},\dfrac{y_C+y_B}{2}\right)=(4,3)[/tex]

[tex]\implies \dfrac{x_C+x_B}{2}=4 \qquad\textsf{and}\qquad \dfrac{y_C+y_B}{2}\right)=3[/tex]

[tex]\implies x_C+x_B=8\qquad\textsf{and}\qquad y_C+y_B=6[/tex]

Since F is the midpoint of AC:

[tex]\left(\dfrac{x_C+x_A}{2},\dfrac{y_C+y_A}{2}\right)=(3,2)[/tex]

[tex]\implies \dfrac{x_C+x_A}{2}=3 \qquad\textsf{and}\qquad \dfrac{y_C+y_A}{2}\right)=2[/tex]

[tex]\implies x_C+x_A=6\qquad\textsf{and}\qquad y_C+y_A=4[/tex]

Add the x-value sums together:

[tex]x_B+x_A+x_C+x_B+x_C+x_A=4+8+6[/tex]

[tex]2x_A+2x_B+2x_C=18[/tex]

[tex]x_A+x_B+x_C=9[/tex]

Substitute the x-coordinate sums found using the midpoint formula into the sum equation, and solve for the x-coordinates of the vertices:

[tex]\textsf{As \;$x_B+x_A=4$, then:}[/tex]

[tex]x_C+4=9\implies x_C=5[/tex]

[tex]\textsf{As \;$x_C+x_B=8$, then:}[/tex]

[tex]x_A+8=9 \implies x_A=1[/tex]

[tex]\textsf{As \;$x_C+x_A=6$, then:}[/tex]

[tex]x_B+6=9\implies x_B=3[/tex]

Add the y-value sums together:

[tex]y_B+y_A+y_C+y_B+y_C+y_A=6+6+4[/tex]

[tex]2y_A+2y_B+2y_C=16[/tex]

[tex]y_A+y_B+y_C=8[/tex]

Substitute the y-coordinate sums found using the midpoint formula into the sum equation, and solve for the y-coordinates of the vertices:

[tex]\textsf{As \;$y_B+y_A=6$, then:}[/tex]

[tex]y_C+6=8\implies y_C=2[/tex]

[tex]\textsf{As \;$y_C+y_B=6$, then:}[/tex]

[tex]y_A+6=8 \implies y_A=2[/tex]

[tex]\textsf{As \;$y_C+y_A=4$, then:}[/tex]

[tex]y_B+4=8\implies y_B=4[/tex]

Therefore, the coordinates of the vertices A, B and C are:

Answer:The interior angle of a polygon is given by

The exterior angle of a polygon is given by

where n is the number of sides of the polygon

The statement

The interior of a regular polygon is 5 times the exterior angle is written as

Solve the equation

That's

Since the denominators are the same we can equate the numerators

That's

180n - 360 = 1800

180n = 1800 + 360

180n = 2160

Divide both sides by 180

n = 12

I).

The interior angle of the polygon is

The answer is

150°

II.

Interior angle + exterior angle = 180

From the question

Interior angle = 150°

So the exterior angle is

Exterior angle = 180 - 150

We have the answer as

30°

III.

The polygon has 12 sides

IV.

The name of the polygon is

Dodecagon

Step-by-step explanation:

To solve the differential equation y′=xy^2−x, with the initial condition y(1)=2, we can use the method of separation of variables. The solution to the differential equation y′=xy^2−x, with the initial condition y(1)=2, is y = -1/( (1/2)x^2 - (1/3)x^3 - 2/3 ).

Step 1: Rewrite the equation in a more convenient form:
y′=xy^2−x

Step 2: Separate the variables by moving all terms involving y to one side and all terms involving x to the other side:
y′ - y^2 = x - x^2

Step 3: Integrate both sides of the equation with respect to x:
∫(1/y^2) dy = ∫(x - x^2) dx

Step 4: Evaluate the integrals:
-1/y = (1/2)x^2 - (1/3)x^3 + C

Step 5: Solve for y by taking the reciprocal of both sides:
y = -1/( (1/2)x^2 - (1/3)x^3 + C )

Step 6: Use the initial condition y(1)=2 to find the value of C:
2 = -1/( (1/2)(1)^2 - (1/3)(1)^3 + C )
2 = -1/(1/2 - 1/3 + C)
2 = -1/(1/6 + C)
2 = -6/(1 + 6C)

Step 7: Solve for C:
1 + 6C = -6/2
1 + 6C = -3
6C = -4
C = -4/6
C = -2/3

Step 8: Substitute the value of C back into the equation for y:
y = -1/( (1/2)x^2 - (1/3)x^3 - 2/3 )

Therefore, the solution to the differential equation y′=xy^2−x, with the initial condition y(1)=2, is y = -1/( (1/2)x^2 - (1/3)x^3 - 2/3 ).

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The correct expansion of the determinant of the given 3x3 matrix is: det [¹2 -4 3 -2 5 0] = 4det + det(1) - 2det [¹2] + 3det [¹] - 2det [¹2 -4 3 -2 5 0].

To expand the determinant of a 3x3 matrix, we use the formula:

det [a b c d e f g h i] = aei + bfg + cdh - ceg - bdi - afh.

For the given matrix [¹2 -4 3 -2 5 0], we can use the above formula to expand the determinant:

det [¹2 -4 3 -2 5 0] = (1)(5)(0) + (2)(-2)(3) + (-4)(-2)(0) - (-4)(5)(3) - (2)(-2)(0) - (1)(-2)(0).

Simplifying this expression gives:

det [¹2 -4 3 -2 5 0] = 0 + (-12) + 0 - (-60) - 0 - 0 = -12 + 60 = 48.

Therefore, the correct expansion of the determinant of the given matrix is: det [¹2 -4 3 -2 5 0] = 4det + det(1) - 2det [¹2] + 3det [¹] - 2det [¹2 -4 3 -2 5 0].

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a) The zero vector: (0, 0, 0)

b) The negative of (2, 1, 3): (-2, -1, -3)

c) The vector c(r, y, z) with c =  and (x, y, z) = (4, 9, 16): (4, 9, 16)

d) The vector (2, 3, 1) + (3, 1, 2): (6, 3, 2)

e) Expressing (1, 4, 32) as a linear combination of p = (1, 2, 2), q = (2, 1, 2), and r = (2, 2, 1):

(1, 4, 32) = (17/7) * (1, 2, 2) + (-70/21) * (2, 1, 2) + (-26/7) * (2, 2, 1).

To find the vectors in P³, we'll use the given operations of vector addition and scalar multiplication.

a) The zero vector:

The zero vector in P³ is the vector where all components are zero. Thus, the zero vector is (0, 0, 0).

b) The negative of (2, 1, 3):

To find the negative of a vector, we simply negate each component. The negative of (2, 1, 3) is (-2, -1, -3).

c) The vector c(r, y, z), where c =  and (x, y, z) = (4, 9, 16):

To compute c(x, y, z), we multiply each component of the vector by the scalar c. In this case, c =  and (x, y, z) = (4, 9, 16). Therefore, c(x, y, z) = ( 4, 9, 16).

d) The vector (2, 3, 1) + (3, 1, 2):

To perform vector addition, we add the corresponding components of the vectors. (2, 3, 1) + (3, 1, 2) = (2 + 3, 3 + 1, 1 + 2) = (5, 4, 3).

e) Expressing (1, 4, 32) as a linear combination of p = (1, 2, 2), q = (2, 1, 2), and r = (2, 2, 1):

To express a vector as a linear combination of other vectors, we need to find scalars a, b, and c such that a * p + b * q + c * r = (1, 4, 32).

Let's solve for a, b, and c:

a * (1, 2, 2) + b * (2, 1, 2) + c * (2, 2, 1) = (1, 4, 32)

This equation can be rewritten as a system of linear equations:

a + 2b + 2c = 1

2a + b + 2c = 4

2a + 2b + c = 32

To solve this system of equations, we can use the method of Gaussian elimination or matrix operations.

Setting up an augmented matrix:

1  2  2  |  1

2  1  2  |  4

2  2  1  |  32

Applying row operations to transform the matrix into row-echelon form:

R2 = R2 - 2R1

R3 = R3 - 2R1

1  2   2  |  1

0 -3  -2  |  2

0 -2  -3  |  30

R3 = R3 - (2/3)R2

1  2   2   |  1

0 -3  -2   |  2

0  0  -7/3 |  26/3

R2 = R2 * (-1/3)

R3 = R3 * (-3/7)

1  2   2   |  1

0  1  2/3  | -2/3

0  0   1   | -26/7

R2 = R2 - (2/3)R3

R1 = R1 - 2R3

R2 = R2 - 2R3

1  2   0   |  79/7

0  1   0   | -70/21

0  0   1   | -26/7

R1 = R1 - 2R2

1  0   0   |  17/7

0  1   0   | -70/21

0  0   1   | -26/7

The system is now in row-echelon form, and we have obtained the values a = 17/7, b = -70/21, and c = -26/7.

Therefore, (1, 4, 32) can be expressed as a linear combination of p, q, and r:

(1, 4, 32) = (17/7) * (1, 2, 2) + (-70/21) * (2, 1, 2) + (-26/7) * (2, 2, 1).

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(a) P(0) = 9000, P(4) ≈ 23051.

(b) The population will reach 18,000 in approximately 5 years.

(a). To find the population at time t=0, we substitute t=0 into the population growth function:

P(0) = 9000(1.3)[tex]^0[/tex] = 9000

To find the population at time t=4, we substitute t=4 into the population growth function:

P(4) = 9000(1.3)[tex]^4[/tex] ≈ 23051

Therefore, the population at time t=0 is 9000 and the population at time t=4 is approximately 23051.

(b). To determine when the population will reach 18,000, we need to solve the equation:

18000 = 9000(1.3)[tex]^t[/tex]

Divide both sides of the equation by 9000:

2 = (1.3)[tex]^t[/tex]

To solve for t, we can take the logarithm of both sides using any base. Let's use the natural logarithm (ln):

ln(2) = ln((1.3)[tex]^t[/tex])

Using the logarithmic property of exponents, we can bring the exponent t down:

ln(2) = t * ln(1.3)

Now, divide both sides of the equation by ln(1.3) to isolate t:

t = ln(2) / ln(1.3) ≈ 5.11

Therefore, the population will reach 18,000 in approximately 5 years.

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Equivalent ratios are those that can be simplified or reduced to the same value. In other words, two ratios are considered equivalent if one can be expressed as a multiple of the other. Some examples of equivalent ratios are 1:2 and 4:8, 3:5 and 12:20, 9:4 and 18:8, etc.

Answer:  16

Step-by-step explanation:

Vertical Angles:When you have 2 intersecting lines the angles across they are equal

65 = 4x + 1                    >Subtract 1 from sides

64 = 4x                         >Divide both sides by 4

x = 16

Answer:

16

Step-by-step explanation:

4x + 1 = 64. Simplify that and you get 16.

With offer 1, you would make $78,216, while with offer 2, you would make $70,354.04. Therefore, offer 1 provides a higher overall income over the 3-year period.

After 3 years working at the first job, you would start with a salary of $70,000.

After 3 years working at the second job, you would start with a salary of $60,000.

Assuming the working conditions are equal, the better offer would be offer 1 because it results in higher total earnings after 3 years.

With offer 1, you would make $78,216, while with offer 2, you would make $70,354.04. Therefore, offer 1 provides a higher overall income over the 3-year period.

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Answer:

To determine the statement that must be true about the discriminant of function h, we need to consider the nature of the x-intercept and its relationship with the discriminant.

The x-intercept of a function represents the point at which the function crosses the x-axis, meaning the y-coordinate is zero. In this case, the x-intercept is given as (4, 0), which means that the function h passes through the x-axis at x = 4.

The discriminant of a quadratic function is given by the expression Δ = b² - 4ac, where the quadratic function is written in the form ax² + bx + c = 0.

Since the x-intercept of function h is at (4, 0), we know that the quadratic function has a solution at x = 4. This means that the discriminant, Δ, must be equal to zero.

Therefore, the correct statement about the discriminant D is:

C. D = 0

Answer:

C. D = 0

Step-by-step explanation:

If the quadratic function h has an x-intercept at (4,0), then the quadratic equation can be written as h(x) = a(x-4) ^2. The discriminant of a quadratic equation is given by the expression b^2 - 4ac. In this case, since the x-intercept is at (4,0), we know that h (4) = 0. Substituting this into the equation for h(x), we get 0 = a (4-4) ^2 = 0. This means that a = 0. Since a is zero, the discriminant of h(x) is also zero. Therefore, statement c. d = 0 must be true about d, the discriminant of function h.

The correct option is III. Using a spring with greater k. Only option III (using a spring with greater k) would cause this system to oscillate with a shorter period.

The period of oscillation of a spring-mass system is given by T = 2π√(m/k), where m is the mass attached to the spring and k is the spring constant. Therefore, any change that affects either m or k will affect the period of oscillation.

I. Increasing m: According to the equation above, an increase in mass will result in an increase in the period of oscillation. This is because a larger mass requires more force to move it, and therefore it will take longer for the spring to complete one cycle of oscillation.

Therefore, increasing m will not cause the system to oscillate with a shorter period. Thus, option I can be eliminated.

II. Increasing A: The amplitude of oscillation is the maximum displacement from equilibrium. It does not affect the period of oscillation directly, but it does affect the maximum velocity and acceleration of the mass during oscillation. As a result, increasing A will not cause the system to oscillate with a shorter period. Thus, option II can also be eliminated.

III. Using a spring with greater k: According to the equation above, an increase in spring constant k will result in a decrease in the period of oscillation. This is because a stiffer spring requires more force to stretch it by a certain amount, resulting in a faster rate of oscillation.

Therefore, using a spring with greater k will cause the system to oscillate with a shorter period.

Therefore, the correct answer is option III.

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