if the answer is true write true if the answer is false replace the underlined word or phrase with the correct forming a hypothesis is the first step of

A government seeking to increase output through the use of fiscal policy may consider various measures, such as increasing government spending, cutting taxes, or a combination of both.

The fiscal policy is the means by which the government controls its spending and tax collection to influence the economy. It is a set of measures adopted by the government to control its spending, tax collection, and borrowing to achieve economic goals. Governments can use fiscal policy to stimulate economic growth, increase output, and maintain price stability.

By increasing government spending, the government can provide a boost to the economy. When the government increases spending, it creates more jobs and income, which in turn, stimulates economic activity. Similarly, tax cuts can stimulate the economy by leaving consumers with more disposable income, which they can use to buy goods and services. This, in turn, increases demand and boosts economic activity.

Thus, governments seeking to increase output through the use of fiscal policy may consider a combination of spending increases and tax cuts.

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The area of the region that is enclosed between y = 11x²-x² + x and y = x² + 25x is -64/3 square units.

The given functions are y = 11x² - x² + x and y = x² + 25x. We need to find the area of the region enclosed by these two functions.What is the region enclosed by two curves?The region enclosed by two curves is defined as the area that lies between two curves. The region may be on either side of the x-axis or the y-axis and also may be closed or open. There are different ways to find the ar the  enclosed by two curves such as using vertical strips or horizontal strips or by integration.Here, we can use the method of integration to find the area of the region enclosed by two curves. We will integrate the difference between the two given functions with respect to x to find the area. Therefore, we get:∫[y = 11x² - x² + x] [y = x² + 25x] dy= ∫[11x² - x² + x] [x² + 25x] dxThe limits of x are given by the points of intersection of the two curves.11x² - x² + x = x² + 25x12x² - 24x = 0x(12x - 24) = 0x = 0, x = 2When x = 0, y = 0 + 0 + 0 = 0When x = 2, y = 11(2²) - 2² + 2 = 24The limits of integration are 0 and 2. Therefore, we get∫[0 to 2] [11x² - x² + x - (x² + 25x)] dx= ∫[0 to 2] [10x² - 24x] dx= 10 ∫[0 to 2] [x² - 2.4x] dx= 10 [(x³/3) - 2.4(x²/2)] [0 to 2]= 10 [(8/3) - 2.4(2) - 0]= 10 [(8/3) - (24/5)]= 10 [(40 - 72)/15]= 10 (-32/15)= -64/3 square unitsTherefore, the area of the region that is enclosed between y = 11x²-x² + x and y = x² + 25x is -64/3 square units.

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The regression equation for the given set of scores, y = 0.5x - 8, is false.

In a regression equation, the slope (denoted as "b") represents the rate of change of the dependent variable (y) with respect to the independent variable (x). In this case, the slope is 0.5, indicating that for every unit increase in x, y increases by 0.5. However, the given information does not provide the value of the slope (b), so we cannot determine its exact value.

To determine if the regression equation y = 0.5x - 8 is true, we need to consider other information such as the correlation coefficient (r) and the standard deviations of x and y. The given information includes the mean of x (mx = 4), the mean of y (my = 10), the sum of squares of x (ssx = 10), and the standard deviation of the predicted scores (sp = 5). However, it does not provide information about the correlation coefficient or the standard deviation of y (sdy).

Without knowledge of the correlation coefficient and the standard deviation of y, we cannot definitively determine the correctness of the regression equation y = 0.5x - 8 based solely on the provided information. Therefore, the answer is b. false.

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Given the differential equation is [tex]- 36 sin(6t) y + 36y = -36 sin (6t)[/tex].We need to find a particular solution to this differential equation. Let's start with the complementary solution where - 36 sin(6t) y + 36y = 0.Using characteristic equation, we have [tex]r² + 36 = 0⇒ r² = - 36⇒ r = ±6i[/tex]

Thus, the complementary solution is given by [tex]y(c) = c1cos(6t) + c2sin(6t)[/tex]Now, we need to find a particular solution to the differential equation. We will assume that the particular solution to be of the form of: yp = A sin (6t) + B cos (6t)Taking first derivative of yp, we haveyp' = 6A cos (6t) - 6B sin (6t)Taking second derivative of yp, we haveyp'' = -36A sin (6t) - 36B cos (6t)Substituting the values of yp, yp' and yp'' in the differential equation[tex],-36 sin(6t) yp + 36yp = -36 sin (6t)⇒ [-36 sin(6t) (A sin (6t) + B cos (6t))] + [36 (A sin (6t) + B cos (6t))] = -36 sin (6t)⇒ (-36A sin² (6t) + 36B cos (6t) sin (6t)) + (36A sin (6t) + 36B cos² (6t)) = -36 sin (6t)⇒ (-36A + 36B) sin (6t) cos (6t) = -36 sin (6t)[/tex]Comparing coefficients of sin (6t) on both sides,

we get:-36A + 36B = 0⇒ A = B On substituting A = B in the above differential equation, we get[tex]:18A sin (6t) cos (6t) = - 18 sin (6t)⇒ A cos (6t) = -1/2Solving for A, we have :A = -1/2cos(6t)[/tex]Thus, the particular solution is given by[tex]: yp = -1/2cos(6t) (sin (6t) + cos (6t))= -1/2 (sin(6t) cos(6t) + cos² (6t))[/tex]Therefore, the general solution is given[tex]by:y(t) = y(c) + yp= c1 cos(6t) + c2 sin(6t) - 1/2 (sin(6t) cos(6t) + cos² (6t))[/tex]Hence, the particular solution to[tex]-36 sin(6t) y + 36y = -36 sin (6t) is -1/2 (sin(6t) cos(6t) + cos² (6t))  y(t) = c1 cos(6t) + c2 sin(6t) - 1/2 (sin(6t) cos(6t) + cos² (6t)).[/tex]

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The derivative of the function is [tex]\frac{d}{dt}\frac{(7t-3)^6}{t+5}[/tex] = [tex]\frac{42(t+5)(7t-3)^5- (7t-3)^6}{(t+5)^2}[/tex]

To find the derivative of the function, follow these steps:

Therefore, the derivative of the expression is [tex]\frac{42(t+5)(7t-3)^5- (7t-3)^6}{(t+5)^2}[/tex]

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Cramer's ruleCramer's rule is a method used to solve a system of linear equations. Consider the following system of linear equations, where a1, b1, c1, a2, b2, and c2 are constants.

These equations can be written in matrix form as follows:

SolutionUsing Cramer's rule, we will first find the determinant of the coefficient matrix (the matrix of the coefficients of the variables):[tex]$$\begin{array}{|cc|} 5 & 2 \\ 7 & 3 \end{array}$$[/tex]

The determinant of the coefficient matrix is 1, so the system has a unique solution.

Now we will find the determinants of the matrices obtained by replacing each column of the coefficient matrix with the column of constants (the values on the right-hand side of the equal sign).[tex]$$\begin{array}{|cc|} 9 & 2 \\ -3 & 3 \end{array}$$[/tex]

The determinant of this matrix is 33.

Now we will find the determinant of the matrix obtained by replacing the second column of the coefficient matrix with the column of constants.[tex]$$ \begin{array}{|cc|} 5 & 9 \\ 7 & -3 \end{array}$$[/tex]

The determinant of this matrix is -44. Using these determinants, we can solve for the variables. [tex]$$x_1=\frac{\begin{array}{|cc|} 9 & 2 \\ -3 & 3 \end{array}}{\begin{array}{|cc|} 5 & 2 \\ 7 & 3 \end{array}}=\frac{33}{1}=33$$[/tex]

Similarly, [tex]$$x_2=\frac{\begin{array}{|cc|} 5 & 9 \\ 7 & -3 \end{array}}{\begin{array}{|cc|} 5 & 2 \\ 7 & 3 \end{array}}=\frac{-44}{1}=-44$$[/tex]

Therefore, the solution to the system of linear equations is x1 = 33 and x2 = -44.

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The conversions of the given lengths are:4.1 × 106 m = 4.1 × 103 km0.01 m = 10 mm4.2 × 10-10 m = 1.13 × 10-7 pm7.6 × 104 m = 7.6 × 102 hm.

The metric prefixes can be used to rewrite the given lengths in meters so that the numerical value is larger than one but less than 1000. The metric prefixes are used to indicate values that are greater or smaller than the unit in use.

Some of the common prefixes used are as follows:

PrefixSymbolMultiple of metermega M106kilok103hectoh102deca10-1deci10-2centi10-3milli10-6micro10-9nano10-12pico10-15femto10-18atto10-21zepto10-24yocto10-27

The given lengths are:4.1 × 106 m = ?0.01 m = ?4.2 × 10-10 m = ?7.6 × 104 m = ?

The conversions are given as follows: -

:4.1 × 106 m = 4.1 × 106 × 10-3 km = 4.1 × 103 km0.01 m = 10-2 m = 10 mm4.2 × 10-10 m = 4.2 × 10-10 × 1012 pm = 4.2 × 2.687 pm = 1.13 × 10-7 pm7.6 × 104 m = 7.6 × 104 × 10-2 hm = 7.6 × 102 hm.

Therefore, the conversions of the given lengths are:4.1 × 106 m = 4.1 × 103 km0.01 m = 10 mm4.2 × 10-10 m = 1.13 × 10-7 pm7.6 × 104 m = 7.6 × 102 hm.

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Given differential equation is `dy/dx = y(x) = 7√x - 3x√x + 6` and the particular point is `(1,4)`To find the particular function y(x) that satisfies the given differential equation, we need to solve the differential equation with the help of integration.

Separate the variables and integrate both sides with respect to x.∫`1/y dy =` ∫`(7√x - 3x√x + 6) dx`Taking integral, we get: `ln|y| + C = 2(√x)^3 - 3(√x)^2 + 6x + C_1` Here, C and C1 are the constants of integration.

Combining both of them we get:C2 = C1 − C where

C2 = constant of integration

= `ln|y|` − `2(√x)^3 + 3(√x)^2 − 6x`

Putting x = 1 and y = 4, we get: `ln(4) - 2 + 3 - 6 = C_2`=> `

ln(4) - 5 = C_2`

So, the required particular function is: Putting the value of `C_2` back into the equation and then simplifying it Taking antilogarithm on both sides Therefore, the particular function is: `y = ± e^(2√x(x + 3) − 5)`

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To find the value of c such that 1+√2 is a root of x²-2 x+c=0 . The value of c that satisfies the condition is c = -1.

1 + √2 is a root of the equation x² - 2x + c = 0, we can use the concept of Vieta's formulas.

Vieta's formulas state that for a quadratic equation of the form ax² + bx + c = 0 with roots α and β, the sum of the roots is given by α + β = -b/a and the product of the roots is given by αβ = c/a.

In this case, we are given that 1 + √2 is a root, so we have α = 1 + √2.

Using Vieta's formulas, we can set up the following equations:

α + β = -(-2)/1

1 + √2 + β = 2

β = 2 - 1 - √2

β = 1 - √2

Now, we can calculate the product of the roots:

αβ = c/1

(1 + √2)(1 - √2) = c/1

1 - 2 = c

c = -1

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Elasticity is defined as the percentage change in quantity demanded of a good or service when there is a change in price.

It is also known as the price elasticity of demand.

[tex]The formula for elasticity is:$$E(x)=-\frac{dQ(x)/dx}{Q(x)/dx}\times\frac{p}{q}$$[/tex]

[tex]The demand function is given by:$$Q(x)=200-8x$$[/tex]

[tex]Differentiating it with respect to x, we get:$$\frac{dQ(x)}{dx}=-8$$[/tex]

[tex]Hence, we have:$$E(x)= -\frac{-8}{200-8x}\times\frac{x}{p}$$$$\implies E(x)=\frac{2}{25-x}$$[/tex]

To find the elasticity, we need to differentiate the demand function, D(x), with respect to x and then multiply it by x divided by D(x).

The demand function is given as:

[tex]q = D(x) = 200 - 8x[/tex]

Differentiating D(x) with respect to x, we get:

[tex]D'(x) = -8[/tex]

To calculate the elasticity, we multiply D'(x) by x and divide it by D(x):

[tex]E(x) = (x * D'(x)) / D(x)[/tex]

Substituting the values we obtained:

[tex]E(x) = (x * (-8)) / (200 - 8x)[/tex]

Simplifying further:

[tex]E(x) = -8x / (200 - 8x)[/tex]

Therefore, the correct option is:

[tex]O D. E(x) = x / (25 - x)[/tex]

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The elasticity can be calculated using the formula below:

(dQ/dx) (P/Q) = E(x) Where P/Q is the ratio of the product's price to the quantity sold, and dQ/dx is the derivative of the quantity function with respect to x.

However, we are unable to determine the elasticity because the product's price is not specified in the problem. Therefore, none of the options are correct.

Alternative E(x)=X 200-8xChoice E(x)=X x -25

Alternative E(x)=x(25-x)

Alternative E(x)=x/(25-x)

E(x) = - (dQ/dx) (P/Q)

Where dQ/dx is the derivative of the quantity function with respect to x, and P/Q is the ratio of the price of the product and the quantity of the product sold.However, since the price of the product is not given in the problem, we cannot calculate the elasticity.

Therefore, the answer is none of the above.

Option E(x)= X 200-8xOption E(x)= X x-25Option E(x)=x(25-x)Option E(x)=x/(25-x)

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The length of the curve is 1 unit.

The given curve is y = 8. We need to find the length of the curve using a table of integrals.

Firstly, let us draw the given curve:

We can see that the curve y = 8 is a horizontal line. As it is a straight line, we can directly find its length using the distance formula.We can find the length of the curve using the distance formula that is given as:

∫[a, b] √(1 + (dy/dx)²) dx where a and b are the limits of the curve and dy/dx is the derivative of the given equation.

For the given curve, y = 8,

dy/dx = 0

Therefore, the length of the curve from x = 0 to x = 1 is given by the distance formula:

∫[0, 1] √(1 + (dy/dx)²) dx= ∫[0, 1] √(1 + 0²) dx

=∫[0, 1] √1 dx= ∫[0, 1] 1

dx= [x] [0,1]=

1 - 0= 1 units

Therefore, the length of the curve is 1 unit.

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To find the determinants, let's calculate them step by step:

(a) Determinant of \( A^T \):

Taking the transpose of \( A \) does not change the determinant. Therefore, \( \left|A^T\right| = \left|A\right| \).

(b) Determinant of \( A^2 \):

To find the determinant of \( A^2 \), we first need to calculate \( A^2 \).

\( A^2 = A \cdot A \)

\( = \begin{bmatrix} 4 & -7 \\ 5 & 8 \end{bmatrix} \cdot \begin{bmatrix} 4 & -7 \\ 5 & 8 \end{bmatrix} \)

\( = \begin{bmatrix} 4 \cdot 4 + (-7) \cdot 5 & 4 \cdot (-7) + (-7) \cdot 8 \\ 5 \cdot 4 + 8 \cdot 5 & 5 \cdot (-7) + 8 \cdot 8 \end{bmatrix} \)

\( = \begin{bmatrix} 16 - 35 & -28 - 56 \\ 20 + 40 & -35 + 64 \end{bmatrix} \)

\( = \begin{bmatrix} -19 & -84 \\ 60 & 29 \end{bmatrix} \)

Now we can find the determinant of \( A^2 \):

\( \left|A^2\right| = \left|-19 \cdot 29 - (-84) \cdot 60\right| \)

\( = \left|-551 - (-5040)\right| \)

\( = \left|-551 + 5040\right| \)

\( = \left|4490\right| \)

\( = 4490 \)

(c) Determinant of \( A \cdot A^T \):

To find the determinant of \( A \cdot A^T \), we need to calculate \( A \cdot A^T \).

\( A \cdot A^T = \begin{bmatrix} 4 & -7 \\ 5 & 8 \end{bmatrix} \cdot \begin{bmatrix} 4 & 5 \\ -7 & 8 \end{bmatrix} \)

\( = \begin{bmatrix} 4 \cdot 4 + (-7) \cdot (-7) & 4 \cdot 5 + (-7) \cdot 8 \\ 5 \cdot 4 + 8 \cdot (-7) & 5 \cdot 5 + 8 \cdot 8 \end{bmatrix} \)

\( = \begin{bmatrix} 16 + 49 & 20 - 56 \\ 20 - 56 & 25 + 64 \end{bmatrix} \)

\( = \begin{bmatrix} 65 & -36 \\ -36 & 89 \end{bmatrix} \)

Now we can find the determinant of \( A \cdot A^T \):

\( \left|A \cdot A^T\right| = \left|65 \cdot 89 - (-36) \cdot (-36)\right| \)

\( = \left|5785 - 1296\right| \)

\( = \left|4489\right| \)

\( = 4489 \)

(d) Determinant of \( 2A \):

To find the determinant of \(

2A \), we multiply each element of \( A \) by 2.

\( 2A = \begin{bmatrix} 2 \cdot 4 & 2 \cdot -7 \\ 2 \cdot 5 & 2 \cdot 8 \end{bmatrix} \)

\( = \begin{bmatrix} 8 & -14 \\ 10 & 16 \end{bmatrix} \)

Now we can find the determinant of \( 2A \):

\( \left|2A\right| = \left|8 \cdot 16 - (-14) \cdot 10\right| \)

\( = \left|128 + 140\right| \)

\( = \left|268\right| \)

\( = 268 \)

(e) Determinant of \( A^{-1} \):

To find the determinant of the inverse of \( A \), we first need to find the inverse of \( A \).

The inverse of \( A \) is given by:

\[ A^{-1} = \frac{1}{\left|A\right|} \cdot \text{adj}(A) \]

where \( \text{adj}(A) \) represents the adjugate of matrix \( A \).

Calculating the determinant of \( A \):

\( \left|A\right| = 4 \cdot 8 - (-7) \cdot 5 \)

\( = 32 + 35 \)

\( = 67 \)

Calculating the adjugate of \( A \):

\[ \text{adj}(A) = \begin{bmatrix} 8 & 7 \\ -5 & 4 \end{bmatrix} \]

Now we can find the determinant of \( A^{-1} \):

\[ \left|A^{-1}\right| = \frac{1}{\left|A\right|} \cdot \left|\text{adj}(A)\right| \]

\[ = \frac{1}{67} \cdot \left|8 \cdot 4 - 7 \cdot (-5)\right| \]

\[ = \frac{1}{67} \cdot \left|32 + 35\right| \]

\[ = \frac{1}{67} \cdot 67 \]

\[ = 1 \]

Therefore, \( \left|A^{T}\right| = \left|A\right| = 67 \), \( \left|A^{2}\right| = 4490 \), \( \left|A A^{T}\right| = 4489 \), \( \left|2 A\right| = 268 \), and \( \left|A^{-1}\right| = 1 \).

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The domain is all real numbers for question (a), (b), (c).  The domain is all real numbers except -3 for question (d). The domain is all real numbers except 0 for question (e) and (f). The y-intercept for (a) is (0, 1). The y-intercept for (b) is (0, 5). The y-intercept for (c) is (0, 0). There is no y-intercept for (d), (e) and (f). The x-intercept for (a) is ( (1/18, 0), for (b) is (5, 0), for (c) is (0, 0), for (d) is (3, 0), for (e) is (1, 0) and (-5, 0) and for (f) is  (2, 0), (3, 0), and (1, 0).

a. The domain of a function represents all possible values of x for which the function is defined. In the case of f(x) = 1 - 18x, there are no restrictions on the values of x. Therefore, the domain is all real numbers.
b. For the function g(x) = 5 - x, again, there are no restrictions on the values of x. Hence, the domain is all real numbers.
c. The function y = 3x is defined for all real numbers since there are no restrictions. Therefore, the domain is all real numbers.
d. In the function h(x) = (3 - x) / (3 + x), the denominator cannot be zero because division by zero is undefined. Setting the denominator equal to zero, we get x = -3. Thus, the domain is all real numbers except -3.
e. For the function y = (x^2 + 4x - 5) / x^2, again, the denominator cannot be zero. Setting the denominator equal to zero, we find x = 0. Therefore, the domain is all real numbers except 0.
f. The function y = t^2 - 5t + (6 / t) is defined for all real numbers except when t = 0, as division by zero is undefined. Thus, the domain is all real numbers except 0.

2. To find the y-intercept, we set x = 0 and evaluate the equation.
a. For f(x) = 1 - 18x, when x = 0, we have f(0) = 1 - 18(0) = 1. Therefore, the y-intercept is (0, 1).
b. For g(x) = 5 - x, when x = 0, we have g(0) = 5 - 0 = 5. Hence, the y-intercept is (0, 5).
c. For y = 3x, when x = 0, we have y = 3(0) = 0. Therefore, the y-intercept is (0, 0).
d. For h(x) = (3 - x) / (3 + x), finding the y-intercept involves setting x = 0. However, in this case, the function is undefined at x = 0. Thus, there is no y-intercept.
e. For y = (x^2 + 4x - 5) / x^2, when x = 0, the function is undefined. Therefore, there is no y-intercept.
f. For y = t^2 - 5t + (6 / t), when t = 0, the function is undefined. Hence, there is no y-intercept.

To find the x-intercept, we set y = 0 and solve for x.
a. For f(x) = 1 - 18x, setting y = 0 gives 1 - 18x = 0. Solving for x, we have x = 1/18. Therefore, the x-intercept is (1/18, 0).
b. For g(x) = 5 - x, when y = 0, we have 5 - x = 0. Solving for x, we find x = 5. Hence, the x-intercept is (5, 0).
c. For y = 3x, setting y = 0 gives 3x = 0. Solving for x, we obtain x = 0. Thus, the x-intercept is (0, 0).
d. For h(x) = (3 - x) / (3 + x), setting y = 0 leads to (3 - x) / (3 + x) = 0. Cross-multiplying, we have 3 - x = 0. Solving for x, we get x = 3. Therefore, the x-intercept is (3, 0).
e. For y = (x^2 + 4x - 5) / x^2, when y = 0, we have (x^2 + 4x - 5) / x^2 = 0. Multiplying both sides by x^2, we obtain x^2 + 4x - 5 = 0. Factoring the quadratic equation, we have (x - 1)(x + 5) = 0. Setting each factor equal to zero, we find x = 1 and x = -5. Hence, the x-intercepts are (1, 0) and (-5, 0).
f. For y = t^2 - 5t + (6 / t), setting y = 0 leads to t^2 - 5t + (6 / t) = 0. Multiplying both sides by t, we have t^3 - 5t^2 + 6 = 0. Factoring the cubic equation, we get (t - 2)(t - 3)(t - 1) = 0. Setting each factor equal to zero, we find t = 2, t = 3, and t = 1. Thus, the x-intercepts are (2, 0), (3, 0), and (1, 0).

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Answer:

  70°

Step-by-step explanation:

You want to know the measure of the second acute angle in a right triangle that has one acute angle measuring 20°.

The angle sum theorem tells you the sum of angles in a triangle is 180°. The square box in the corner of this triangle tells you that is a right angle that has a measure of 90°. Then the sum of angles is ...

  20° +90° +x = 180°

  x = 180° -110° = 70°

The measure of unknown angle x is 70°.

__

Additional comment

Since the measure of a right angle is always 90°, the sum of the measures of the two acute angles in a right triangle is 180°-90° = 90°. That is, they are complementary.

  x = 90° -20° = 70°

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The formula to use to write sin(10.6v) - sin(3.4v) as a product is: B. difference of sines: [tex]2cos(\frac{x\;+\;y}{2} )sin(\frac{x\;-\;y}{2} )[/tex].

The average of the original inputs is [tex]\frac{x\;+\;y}{2} =7v[/tex]

Half the distance between the original inputs is [tex]\frac{x\;-\;y}{2} =3.6v[/tex]

The difference as a product is: sin(10.6v) - sin(3.4v) = 2cos(7v)sin(3.6v).

In Mathematics and Geometry, the Bhaskaracharya sum and difference formulas that shows the relationship between sine and cosine values for trigonometric identities (two angles) can be modeled by the following mathematical equation;

sin(u + v) = sin(u)cos(v) + cos(u)sin(v)

sin(u - v) = sin(u)cos(v) - cos(u)sin(v)

sin(u + v) - sin(u - v) = 2cos(u)sin(v)

In this context, the formula to use in writing cos(17.6p) + cos(8.4p) as a product is given by:

sum of cosines: [tex]2cos(\frac{x\;+\;y}{2} )sin(\frac{x\;-\;y}{2} )[/tex].

For the average of the original inputs, we have:

[tex]\frac{x\;+\;y}{2} =\frac{10.6\;+\;3.4}{2} \\\\\frac{x\;+\;y}{2} =7v[/tex]

For half the distance between the original inputs, we have:

[tex]\frac{x\;-\;y}{2} =\frac{10.6\;-\;3.4}{2} \\\\\frac{x\;-\;y}{2} =3.6v[/tex]

Therefore, the sum as a product can be written as follows:

sin(10.6v) - sin(3.4v) = 2cos(7v)sin(3.6v).

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The characteristic equation is given by [tex]r² - 2r - 3 = 0[/tex] ⇒ [tex](r - 3)(r + 1) = 0[/tex] ⇒ [tex]r = 3[/tex] and [tex]r = -1[/tex]. The complementary function is [tex]yc = c₁e³x + c₂e⁻x[/tex]  where c₁ and c₂ are constants.

To identify the particular integral, let's guess that it has the form:

[tex]yp = Axex[/tex]

Since the function is already present in the complementary function and the differential equation is not homogeneous, let's guess the particular integral takes the form:

[tex]yp = A x²ex[/tex]

Therefore,

[tex]yp' = (2Ax + A)ex[/tex]

[tex]yp'' = (4A + 2A)ex[/tex]

[tex]= 6Aex[/tex]

Substituting these into the differential equation:

[tex]6Aex − 2(2Ax + A)ex − 3A x²ex = 4ex − 6[/tex]

==> [tex]2Axex = 4ex[/tex]

==> [tex]A = 2[/tex]

Therefore, the particular integral is given by:

[tex]yp = 2 x²ex[/tex]

The general result of the differential equation:

[tex]y = yc + yp[/tex]

[tex]= c₁e³x + c₂e⁻x + 2 x²ex[/tex]

Hence, the result of the differential equation is given by

[tex]yc = c₁e³x + c₂e⁻x[/tex]

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The function f(x) = {x^2 - 1, 10x - 26, x ≤ c, c < x} is continuous on the interval (-∞, ∞) when c = (1 + √5) / 2.

To determine the value of c for which the function f(x) is continuous on the interval (-∞, ∞), we need to check the conditions at the point where the pieces of the function meet, which is at x = c.

For the function to be continuous at x = c, the left-hand limit as x approaches c should be equal to the right-hand limit as x approaches c, and both limits should be equal to the value of f(x) at x = c.

Let's evaluate the left-hand limit and right-hand limit separately:

Left-hand limit:

lim(x→c-) f(x) = lim(x→c-) (x^2 - 1)

= c^2 - 1

Right-hand limit:

lim(x→c+) f(x) = lim(x→c+) (10x - 26)

= 10c - 26

For the function to be continuous at x = c, the left-hand limit and right-hand limit should be equal to f(c), which is given by f(c) = c.

Therefore, we have the following conditions:

c^2 - 1 = c (1)

10c - 26 = c (2)

Solving equation (1), we have:

c^2 - c - 1 = 0

Using the quadratic formula, we find the roots of this equation:

c = (-(-1) ± √((-1)^2 - 4(1)(-1))) / (2(1))

c = (1 ± √(1 + 4)) / 2

c = (1 ± √5) / 2

Since we are looking for the value of c that makes the function continuous on the interval (-∞, ∞), we need c to be greater than -1. Therefore, we choose the positive root:

c = (1 + √5) / 2

The function f(x) = {x^2 - 1, 10x - 26, x ≤ c, c < x} is continuous on the interval (-∞, ∞) when c = (1 + √5) / 2.

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To show that sin(π/2-θ) = cosθ using an angle difference identity, we can use the following identity: sin(A-B) = sinAcosB - cosAsinB.

Let's substitute A = π/2 and B = θ into the identity: sin(π/2-θ) = sin(π/2)cos(θ) - cos(π/2)sin(θ).

Since sin(π/2) = 1 and cos(π/2) = 0, we can simplify the equation to: sin(π/2-θ) = 1cos(θ) - 0sin(θ).

This simplifies to: sin(π/2-θ) = cos(θ).

Therefore, sin(π/2-θ) = cosθ, as required.

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*Complete question

Which angle difference identity can be used to prove that sin(π/2-θ) = cosθ?

The Fourier series of a periodic square wave is [tex]$f(t) = \frac{\pi}{4} \left[ \sin(\omega t) + \frac{3}{1}\sin(3\omega t) + \frac{5}{1}\sin(5\omega t) + \cdots \right]$ where $f(t)=+1$ for $0<\omega t<\pi$, $2\pi<\omega t<3\pi$[/tex], and so on for [tex]$\pi<\omega t<2\pi, 3\pi<\omega t<4\pi$[/tex], and so on.

Consider a periodic square wave function [tex]$f(t)$[/tex] which takes the value of +1 or -1 in alternate intervals of [tex]$\pi$[/tex].

The Fourier series of a periodic square wave is [tex]$f(t) = \frac{\pi}{4} \left[ \sin(\omega t) + \frac{3}{1}\sin(3\omega t) + \frac{5}{1}\sin(5\omega t) + \cdots \right]$Where $f(t)=+1$ for $0<\omega t<\pi$, $2\pi<\omega t<3\pi$, and so on for $\pi<\omega t<2\pi$, $3\pi<\omega t<4\pi$[/tex], and so on.

In the Fourier series, the coefficients of [tex]$sin(\omega t), sin(3\omega t), sin(5\omega t)$[/tex], and so on are all equal to the half of the amplitude of the waveform.

So, the Fourier series of [tex]$f(t)$[/tex] becomes

[tex]$f(t) = \frac{a_{0}}{2} + \sum_{n=1}^{\infty} \left[ a_{n} \sin(n\omega t) + b_{n} \cos(n\omega t) \right]$[/tex]

In the above expression, [tex]$a_{0}$[/tex] is the DC component, and [tex]$a_{n}$[/tex] and [tex]$b_{n}$[/tex] are the coefficients of sine and cosine terms respectively.

The waveform is an odd function of t, and thus all the cosine terms will be zero.

In order to find the coefficient of [tex]$a_{n}$[/tex], we must compute it as follows:

[tex]$a_{n} = \frac{2}{T}\int_{T}f(t) \sin(n\omega t) dt$[/tex]

Since the wave function is defined for the interval [tex]$T = 2\pi$[/tex], the expression for [tex]$a_{n}$[/tex] becomes

[tex]$a_{n} = \frac{2}{2\pi}\int_{0}^{2\pi}f(t) \sin(n\omega t) dt$[/tex]

The expression for [tex]$a_{n}$[/tex] is obtained by breaking up the waveform into a set of cosine terms. We will get the result

[tex]$a_{n} = \frac{2}{T} \int_{0}^{T} f(t) \sin(n\omega t) dt$[/tex]

Substituting the function, we get, [tex]$a_{n} = \frac{2}{2\pi} \int_{0}^{2\pi} f(t) \sin(n\omega t) dt$[/tex]

Since the function [tex]$f(t)$[/tex] takes the value of +1 for alternate intervals of [tex]$\pi$[/tex], we can replace it in the expression for [tex]$a_{n}$[/tex] as follows:

[tex]$a_{n} = \frac{2}{2\pi} \int_{0}^{\pi} \sin(n\omega t) dt + \frac{2}{2\pi} \int_{\pi}^{2\pi} (-1) \sin(n\omega t) dt$[/tex]

Using the formula [tex]$\int \sin(x) dx = -\cos(x) + C$, we get$$a_{n} = \frac{2}{2\pi} \left[ \frac{-\cos(n\omega t)}{n\omega} \right]_{0}^{\pi} + \frac{2}{2\pi} \left[ \frac{\cos(n\omega t)}{n\omega} \right]_{\pi}^{2\pi}$$[/tex]

Simplifying the expression, we get [tex]$$a_{n} = \frac{2}{n\pi} \left[ 1 - (-1)^{n} \right]$$[/tex]

Thus, the Fourier series for a periodic square wave function [tex]$f(t)$ is$$f(t) = \frac{\pi}{4} \left[ \sin(\omega t) + \frac{3}{1}\sin(3\omega t) + \frac{5}{1}\sin(5\omega t) + \cdots \right]$$[/tex]

The Fourier series of a periodic square wave is [tex]$f(t) = \frac{\pi}{4} \left[ \sin(\omega t) + \frac{3}{1}\sin(3\omega t) + \frac{5}{1}\sin(5\omega t) + \cdots \right]$ where $f(t)=+1$ for $0<\omega t<\pi$, $2\pi<\omega t<3\pi$[/tex], and so on for [tex]$\pi<\omega t<2\pi, 3\pi<\omega t<4\pi$[/tex], and so on. The coefficients of [tex]$sin(\omega t), sin(3\omega t), sin(5\omega t)$[/tex], and so on are all equal to the half of the amplitude of the waveform.

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The distance between (10, 20, 10) and (-12, 6, 12) is √684.

The distance between two points in three-dimensional space can be calculated using the distance formula. Given the points (10, 20, 10) and (-12, 6, 12), the distance between them is determined by the formula:

Distance = √[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]

Substituting the values, we get:

Distance = √[(-12 - 10)^2 + (6 - 20)^2 + (12 - 10)^2]

= √[(-22)^2 + (-14)^2 + (2)^2]

= √[484 + 196 + 4]

= √684

Therefore, the distance between the given points is √684.

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By iteration, g(10) = 60 for (i) and g(10) = 14 for (ii)

Given information: The recursive formula are(i)  t(n) = t(n - 2) + 2n, t(1) = 1, t(0) = 0(ii) t(n) = t(n - 1) + n/2, t(1) = 1.Solve by iteration:

1. The recursive formula is t(n) = t(n - 2) + 2n, t(1)

                                                   = 1, t(0)

                                                   = 0.

This can be written as, t(n) = t(n - 2) + 2nOr, t(n + 2) = t(n) + 2(n + 2) (Add 2(n + 2) to both sides)

Let, the initial value of t(n) be g(n)

Therefore, t(n + 2) = g(n + 2)g(n) + 2(n + 2)

This iteration will be used until the value converges to the answer.

2. The recursive formula is t(n) = t(n - 1) + n/2, t(1) = 1.

This can be written as, t(n) = t(n - 1) + n/2Or, t(n + 1) = t(n) + (n + 1)/2

Let, the initial value of t(n) be g(n)

Therefore, t(n + 1) = g(n + 1)g(n) + (n + 1)/2

This iteration will be used until the value converges to the answer.

3. t(0) = 0 and t(1) = 1.This is given.

The iterative method is used when no exact formula can be found that will provide the n-th term of the sequence directly.

The iteration for

(i) becomes, g(0) = 0,

g(1) = 1.g(2)

     = g(0) + 2(2)

     = 4g(3)

     = g(1) + 2(3)

     = 7g(4)

     = g(2) + 2(4)

     = 12g(5)

     = g(3) + 2(5)

     = 17g(6)

     = g(4) + 2(6)

     = 24g(7)

     = g(5) + 2(7)

     = 31g(8)

     = g(6) + 2(8)

     = 40g(9)

     = g(7) + 2(9)

     = 49g(10)

     = g(8) + 2(10)

     = 60

The iteration for

(ii) becomes, g(1) = 1.g(2)

                           = g(1) + (2)/2

                           = 2g(3)

                           = g(2) + (3)/2

                           = 3.5g(4)

                           = g(3) + (4)/2

                           = 5g(5)

                           = g(4) + (5)/2

                           = 6.5g(6)

                           = g(5) + (6)/2

                           = 8g(7)

                           = g(6) + (7)/2

                           = 9.5g(8)

                          = g(7) + (8)/2

                          = 11g(9)

                          = g(8) + (9)/2

                          = 12.5g(10)

                          = g(9) + (10)/2

                          = 14.

The values of g(n) converge to the corresponding values of t(n).

Therefore, the required answers are: g(10) = 60 for (i) and g(10) = 14 for (ii).

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The partial derivative of f(x, y) with respect to x is[tex]fx = (1/xy) * (x(dy/dx) + y) - 2xye².[/tex]

To find the partial derivative of the function [tex]f(x, y) = ln(xy) - e²xy + y²[/tex] with respect to x, we treat y as a constant and differentiate the expression with respect to x.

The derivative of ln(xy) with respect to x can be found using the chain rule. Let's break it down:

[tex]d/dx [ln(xy)] = (1/xy) * (d/dx [xy])\\= (1/xy) * (x(dy/dx) + y)[/tex]

The derivative of the remaining terms:

d/dx [-e²xy] = -2ye²xy

d/dx [y²] = 0 (since y² is a constant with respect to x)

Now, we can combine the derivatives:

[tex]fx = (1/xy) * (x(dy/dx) + y) - 2ye²xy + 0\\= (1/xy) * (x(dy/dx) + y) - 2xye²[/tex]

The correct partial derivative of f(x, y) with respect to x is[tex]fx = (1/xy) * (x(dy/dx) + y) - 2xye².[/tex]

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The value of `dy/dx` for the given problem is `(8x * sin(x+y)) / (5cos(x+y) - 8x)` and the value of `y` is `-x - (ln|x + 1| / 2) + ln3`.

Given the function `y = x² In(x + 1)` and the equation `5 cos(x+y)-3e = 4x²`, we are to find `dy/dx` by using implicit differentiation.

Using the Chain Rule, we can find `dy/dx` of `y = x² ln(x + 1)` as:

dy/dx = d/dx [x² ln(x + 1)] = (d/dx [x²])(ln(x + 1)) + (x²)(d/dx [ln(x + 1)])

Using the Product Rule, we can find

`d/dx [ln(x + 1)]` as:d/dx [ln(x + 1)] = 1 / (x + 1)

Thus, dy/dx = (2x ln(x + 1)) + (x² / (x + 1)) ............(1)

Now, using the given equation `5 cos(x+y) - 3e = 4x²`, we have to find `dy/dx`.

Differentiating the given equation implicitly with respect to `x`, we get:

5(-sin(x+y))(1 + dy/dx) - 3(0) = 8x`dy/dx`

Rearranging and solving for `dy/dx`, we get:

`dy/dx = (8x * sin(x+y)) / (5cos(x+y) - 8x)`

Substitute this value of `dy/dx` in equation (1), we get:

`(2x ln(x + 1)) + (x² / (x + 1)) = (8x * sin(x+y)) / (5cos(x+y) - 8x)`

Solving this for `y`, we get:y = -x - (ln|x + 1| / 2) + c

Substituting `y = -x - (ln|x + 1| / 2) + c` in the given equation

`5 cos(x+y) - 3e = 4x²`, we get:5 cos(x - (ln|x + 1| / 2) + c) - 3e = 4x²

The value of `c` can be found by using the given point `(5, 0)`.

Substituting these values in the above equation, we get: c = ln3

Now the value of `c` can be substituted to find the value of `y`.

Thus, the solution is `y = -x - (ln|x + 1| / 2) + ln3`.

Therefore, the value of `dy/dx` for the given problem is `(8x * sin(x+y)) / (5cos(x+y) - 8x)` and the value of `y` is `-x - (ln|x + 1| / 2) + ln3`.

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Another vector that has the same projection onto v=〈1, 1〉 as u=〈1, 2〉 is <1/2 - (3/2√2), 3/2 - (3/2√2)> .Below is the required picture of the projection of u onto v=〈1, 1〉

Given a fixed vector v, there is an infinite set of vectors u with the same value of proj vu as below:

u=proj v u+orthogonal component of u with respect to vSince the given vector u = <1,2> has the same projection onto v= <1,1> as u, the scalar projection can be found as:

proj vu = (u⋅v) / ||v||2u = proj v u + orthogonal component of u with respect to v

⇒ proj v u = u − orthogonal component of u with respect to v

Let's find the orthogonal component of u with respect to v.First, we need to find the projection of u onto v, which is,

proj v u = (u⋅v) / ||v||2

= (<1,2> ⋅ <1,1>) / (√2)2

= 3 / 2√2

Then, the orthogonal component of u with respect to v is:

ortho v u

= u − proj v u v

= <1,1>ortho v u

= u − proj v u v

= <1,2> − 3 / 2√2<1,1>

= <1,2> − 3 / 2√2<1,1>

= <1/2 - (3/2√2),

3/2 - (3/2√2)>

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The first derivative of the implicit function is [tex]y' = \left(\frac{\sqrt[6]{e^{4}-x^{6}}}{x^{5}} \right)^{5}[/tex].

In this problem we need to determine the first derivative of an implicit function, that is, a function where variables are not functions of another one. This can be done by means of differentiation rules. First, write the entire expression:

㏑(x⁶ + y⁶) = 4

Second, use differentiation rules:

(6 · x⁵ + 6 · y⁵ · y') / (x⁶ + y⁶) = 0

x⁵ + y⁵ · y' = 0

y' = - y⁵ / x⁵

y' = - (y / x)⁵

Third, eliminate y in the previous result:

㏑(x⁶ + y⁶) = 4

x⁶ + y⁶ = e⁴

y⁶ = e⁴ - x⁶

[tex]y = \sqrt[6]{e^{4}-x^{6}}[/tex]

[tex]y' = \left(\frac{\sqrt[6]{e^{4}-x^{6}}}{x^{5}} \right)^{5}[/tex]

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The cost of selling 300 cellphones is $22,800.3 and the profit earned from selling 300 cellphones is $56,400.

Given, Revenue of selling x cellphones = R(x) = -1.5x² + 314x

The cost of selling x cellphones in dollars is C(x).

We can calculate the cost from revenue by subtracting the profit from the revenue earned.

Here, Profit = Revenue - Cost

Profit earned from selling x cellphones in dollars = R(x) - C(x)

The question asks us to find the profit earned from selling 300 cellphones. Therefore, we need to calculate the revenue, cost, and profit earned from selling 300 cellphones.

1) Revenue earned from selling 300 cellphones

.R(x) = -1.5x² + 314x, x = 300 (given)

R(300) = -1.5(300)² + 314(300)

           = -1.5(90000) + 94200

           = $79,200

Therefore, the revenue earned from selling 300 cellphones is $79,200.2

Cost of selling 300 cellphones.C(x) = 12x + 18,000, x = 300 (given)

C(300) = 12(300) + 18,000

            = $22,800

Therefore, the cost of selling 300 cellphones is $22,800.3

Profit earned from selling 300 cellphones.

Profit = Revenue - Cost

         = R(300) - C(300)

         = $79,200 - $22,800

         = $56,400

Therefore, the profit earned from selling 300 cellphones is $56,400.

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List A consists of the five numbers x, x + 1, x + 1, x + 1, x + 1, and the question requires the determination of how much greater is the arithmetic mean of the numbers in list B than the average of the numbers in list A.

In list B, the numbers are x, x + 2, x + 2, x + 2, x + 2.

The arithmetic mean of numbers in list A is given by the sum of numbers in list

A divided by the number of elements in the list:

Mean of list

A = (x + (x + 1) + (x + 1) + (x + 1) + (x + 1)) / 5

Mean of list

A = (5x + 5) / 5

Mean of list

A = x + 1

The arithmetic mean of numbers in list B is given by the sum of numbers in list B divided by the number of elements in the list:

Mean of list

B = (x + (x + 2) + (x + 2) + (x + 2) + (x + 2)) / 5

Mean of list

B = (5x + 10) / 5

Mean of list

B = x + 2

The difference between the mean of list B and list A is:

Mean of list B - Mean of list

A= (x + 2) - (x + 1)= x + 1

but it is greater than zero.

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The measure of angle A in the isosceles triangle ABC is 62.5 degrees.

In an isosceles triangle ABC, where AB = AC, we are given that angle B (denoted as ∠B) measures 55 degrees. We need to calculate the measure of angle A (denoted as ∠A).

Since AB = AC, we know that angles A and C are congruent (denoted as ∠A ≅ ∠C). In an isosceles triangle, the base angles (the angles opposite the equal sides) are congruent.

Therefore, we have:

∠A ≅ ∠C

Also, the sum of the angles in a triangle is always 180 degrees. Hence, we can write:

∠A + ∠B + ∠C = 180

Substituting the given values:

∠A + 55 + ∠A = 180

Combining like terms:

2∠A + 55 = 180

Subtracting 55 from both sides:

2∠A = 180 - 55

2∠A = 125

Dividing by 2:

∠A = 125 / 2

∠A = 62.5

Therefore, the measure of angle A in the isosceles triangle ABC is 62.5 degrees.

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