A squre is cut into three rectangles X, Y and Z

We can use similar methods to rewrite the expressions in problems 19 through 22 as single power series.

To rewrite each of the expressions in problems 18 through 22 as a single power series whose generic term involves xn, we can use the formula for a geometric series. This formula states that a geometric series with first term a and common ratio r can be expressed as:

a + ar + ar² + ar³ + ...

We can rewrite this formula in terms of xn by letting a = f(21) and r = (x - 21)/21. Then, the nth term of the series is given by:

f(21) ˣ [(x - 21)/21]^n

Using this formula, we can rewrite each of the given expressions as a single power series whose generic term involves xn.

For example, in problem 18, we are given the expression:

f(x) = 1/(x² - 4x + 3)

We can factor the denominator as (x - 1)(x - 3) and write:

f(x) = 1/[(x - 1)(x - 3)]

Using partial fractions, we can express f(x) as:

f(x) = 1/(2(x - 1)) - 1/(2(x - 3))

Now, we can use the formula for a geometric series with a = 1/2, r = (x - 21)/21, and n = k - 21 to write:

f(x) = 1/2 ˣ [(x - 21)/21]²¹ - 1/2 * [(x - 21)/21]²³ + ...

This is a single power series whose generic term involves xn. We can use similar methods to rewrite the expressions in problems 19 through 22 as single power series.

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These are the z-scores that correspond to the 0.005 area in each tail. If a test statistic falls beyond these boundaries (either below -2.576 or above 2.576), it would be in the critical region, and you would reject the null hypothesis.

The critical region, tails, and z-score are essential concepts in hypothesis testing. Let's explore how these terms relate to the given alpha level (α = 0.01).

The critical region is the area in the tails of a distribution where we reject the null hypothesis if the test statistic falls within this region. In other words, it's the area where the probability of finding the test statistic is very low if the null hypothesis were true.

Tails refer to the extreme ends of a distribution. In a standard normal distribution, tails are the areas to the left and right of the main portion of the curve.

The z-score (or standard score) is a measure that expresses the distance of a data point from the mean in terms of standard deviations.

For an alpha level (α) of 0.01, you want to find the z-score boundaries that correspond to the critical region. In this case, the critical region will be in both tails, with a total area of 0.01. Since there are two tails, each tail will contain an area of 0.005 (0.01 / 2).

Using a z-score table or calculator, you can find the z-score boundaries for α = 0.01 as:

z = 2.576 and z = -2.576

These are the z-scores that correspond to the 0.005 area in each tail. If a test statistic falls beyond these boundaries (either below -2.576 or above 2.576), it would be in the critical region, and you would reject the null hypothesis.

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The correct answer is (b) fraction numerator cos squared x over denominator 2 end fraction minus fraction numerator cos cubed x over denominator 3 end fraction plus C.

To evaluate the integral, we can use the identity sin²(x) + cos²(x) = 1 to write sin³(x) = sin²(x) × sin(x) = (1 - cos²(x)) × sin(x). Then, we can use u-substitution with u = cos(x) and du = -sin(x) dx to get:

integral sin³(x) cos²(x) dx = -integral (1 - u²) u² du
= integral (u⁴ - u²) du
= (1/5)u⁵ - (1/3)u³ + C
= (1/5)cos⁵(x) - (1/3)cos³(x) + C

Using the identity cos²(x) = 1 - sin²(x), we can also write the answer as:

(1/2)cos²(x) - (1/3)cos³(x) + C

Therefore, the correct answer is (b) fraction numerator cos squared x over denominator 2 end fraction minus fraction numerator cos cubed x over denominator 3 end fraction plus C.

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The orthogonal projection of u along v is ⟨1, 2, 3⟩.

To find the orthogonal projection of vector u along vector v, you'll need to use the following formula:

Orthogonal projection of u onto [tex]v = (u ⋅ v / ||v||^2) * v[/tex]

Given u = ⟨1, 5, 1⟩ and v = ⟨1, 2, 3⟩, let's calculate the required values.

1. Compute the dot product (u ⋅ v):
u ⋅ v = (1)(1) + (5)(2) + (1)(3) = 1 + 10 + 3 = 14

2. Calculate the magnitude squared of [tex]v (||v||^2):[/tex]
[tex]||v||^2 = (1)^2 + (2)^2 + (3)^2 = 1 + 4 + 9 = 14[/tex]
3. Find the orthogonal projection:
Orthogonal projection = [tex](u ⋅ v / ||v||^2) * v = (14 / 14) * ⟨1, 2, 3⟩ = ⟨1, 2, 3⟩[/tex]

So, the orthogonal projection of u along v is ⟨1, 2, 3⟩.

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Using a t-table or calculator, the t-distribution for an upper tail area of 0.025 with 15 degrees of freedom. The final answer (a) -2.021 and 2.021 (b) -1.675 (c) 0.873, (d) -2.845 and 2.845. (e) -2.021 and 2.021,

a. Using a t-table or calculator, the t-distribution for an upper tail area of 0.025 with 15 degrees of freedom is 2.131. T-values, where 95% of the area falls between, are -2.021 and 2.021.

b. Using a t-table or calculator, the t-value for a lower tail area of 0.05 with 55 degrees of freedom is -1.675.

c. Using a t-table or calculator, the t-value for an upper tail area of 0.20 with 35 degrees of freedom is 0.873.

d. The t-values corresponding to 0.01 and 0.99 quantiles for a t-distribution with 20 degrees of freedom are -2.845 and 2.845, respectively. Therefore, the t-values where 98% of the area falls between are -2.845 and 2.845.

e. The t-values corresponding to 0.025 and 0.975 quantiles for a t-distribution with 40 degrees of freedom are -2.021 and 2.021, respectively.

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After simplification, the exact value of the expression "4/5 × 15/8 × 14/5" is 21/5.

To find the simplified-value of the expression, we simply multiply the numerators and denominators of the fractions in the order given, and then simplify the resulting fraction:

The expression given for simplification is : 4/5 × 15/8 × 14/5;

So, We first write and multiply all the numerators and denominators together,

We get,

4/5 × 15/8 × 14/5 = (4×15×14)/(5×8×5),

⇒ 840/200,

Now, we simplify the above fraction,

We get,

= 21/5.

Therefore, the simplified value is 21/5.

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The given question is incomplete, the complete question is

Find the exact value of the expression "4/5 × 15/8 × 14/5".

A. The expression for the fish population P(t) at time t in years is:
P(t) = 20,000 / (1 + (25 - 1) * e^(-ln(2)t))

B. It will take approximately 7.637 years for the fish population to reach 10,000.

(A) To find an expression for the size of the fish population P at time t in years, we can use the logistic model:
P(t) = K / (1 + e^(-kt))

where P(t) is the population size at time t, K is the carrying capacity (in this case, K = 20000), and k is a constant representing the rate of growth.

To find k, we can use the fact that the number of fish doubled in the first year:
800 * 2 = 1600

So the initial population size P(0) is 800, and the population size after one year (t=1) is 1600. Substituting these values into the logistic model, we get:
1600 = 20000 / (1 + e^(-k*1))

Solving for k, we get:
k = 1.098612
Rounding to six decimal places, k ≈ 1.098612.

Now we can plug in this value of k and the given value of K to get the expression for P(t):
P(t) = 20000 / (1 + e^(-1.098612t))

(B) To find out how long it will take for the fish population to reach 10000, we can set P(t) equal to 10000 and solve for t:
10000 = 20000 / (1 + e^(-1.098612t))

Multiplying both sides by the denominator and simplifying, we get:
1 + e^(-1.098612t) = 2

Subtracting 1 from both sides, we get:
e^(-1.098612t) = 1

Taking the natural logarithm of both sides, we get:
ln(e^(-1.098612t)) = ln(1)

Simplifying, we get:
-1.098612t = 0

Dividing both sides by -1.098612, we get:
t ≈ 0.693

Rounding to three decimal places, it will take approximately 0.693 years (or about 8 months) for the fish population to reach 10000.

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The critical point of f(x) in the interval [0,2π] is x = π/4.

To find the critical points of a function, we need to find the values of x where the derivative of the function is zero or undefined.

First, let's find the derivative of f(x):

f'(x) = 5(cos^2(x) - sin^2(x))

Next, we need to find the values of x where f'(x) = 0 or is undefined.

Setting f'(x) = 0:

5(cos^2(x) - sin^2(x)) = 0

cos^2(x) - sin^2(x) = 0

Using the identity cos^2(x) - sin^2(x) = cos(2x), we get:

cos(2x) = 0

This means that 2x = π/2 or 2x = 3π/2, since the cosine function is zero at these angles.

Solving for x, we get:

x = π/4 or x = 3π/4

However, we need to check if these points are in the interval [0,2π]. Only x = π/4 is in this interval.

Therefore, the critical point of f(x) in the interval [0,2π] is x = π/4.

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To provide an answer, I would need the specific mathematical problem or equation you are trying to solve.

To find the general solution without the use of a calculator or computer, you will need to rely on your knowledge of mathematical concepts and equations. Start by identifying the type of equation you are working with and any relevant formulas that can help you simplify it.

From there, you can use algebraic manipulation to isolate the variable and solve for its possible values. It's important to note that finding the general solution in this way may require a bit of trial and error, so be prepared to test out different approaches until you arrive at the correct answer. With patience and persistence, you can find the solution to many mathematical problems without the aid of a calculator or computer.

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The probability that the next person will purchase no costume is 36% to the nearest whole number.

We have,

To express the probability that the next person will purchase no costume as a percent to the nearest whole number, we need to use the total number of visitors to the website as the denominator and the number of visitors who purchased no costume as the numerator.

The total number of visitors to the website is:

168 + 252 + 47 = 467

The number of visitors who purchased no costume is 168.

So the probability that the next person will purchase no costume is:

168/467 = 0.3609

To express this as a percent, we multiply by 100:

0.3609 × 100

= 36.09

Rounding this to the nearest whole number, we get:

36%

Therefore,

The probability that the next person will purchase no costume is 36% to the nearest whole number.

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Answer:

Y>0: X= -2

Y<0: X= 1

Y=0: X= -1

Step-by-step explanation:

The trigonometric ratio values for the mentioned angles are:

To find the trigonometric ratio values, we use the unit circle which represents the values of sine, cosine, and tangent of all angles in the first quadrant (0 to 90 degrees). From there, we can use reference angles and the periodicity of trigonometric functions to find the values for other angles.

For example, to find Sin 120 degrees, we can use the reference angle of 60 degrees (180 - 120) and the fact that the sine function is negative in the second quadrant, so:

Sin 120 degrees = - Sin 60 degrees = [tex]$-\frac{\sqrt{3}}{2}$[/tex]

Similarly, to find Cos 150 degrees, we can use the reference angle of 30 degrees (180 - 150) and the fact that the cosine function is negative in the third quadrant, so:

Cos 150 degrees = - Cos 30 degrees =[tex]$-\frac{\sqrt{3}}{2}[/tex]

And to find Tan 225 degrees, we can use the reference angle of 45 degrees (225 - 180) and the fact that the tangent function is positive in the second and fourth quadrants, so:

Tan 225 degrees = Tan 45 degrees = 1.

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The Taylor polynomials T2(x) and T3(x) for f(x) = sin(x) centered at x = 0 are given by T2(x) = x and T3(x) = x - x^3/6.

The Taylor polynomials for a function f(x) centered at x = a are given by

Tn(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + ... + f^(n)(a)(x-a)^n/n!

where f^(n)(a) represents the nth derivative of f(x) evaluated at x = a.

For f(x) = sin(x), we have

f(0) = sin(0) = 0

f'(x) = cos(x)

f'(0) = cos(0) = 1

f''(x) = -sin(x)

f''(0) = -sin(0) = 0

f'''(x) = -cos(x)

f'''(0) = -cos(0) = -1

Using these derivatives, we can calculate the Taylor polynomials

T2(x) = f(0) + f'(0)x/1! + f''(0)x^2/2!

= 0 + x/1! + 0x^2/2!

= x

T3(x) = f(0) + f'(0)x/1! + f''(0)x^2/2! + f'''(0)x^3/3!

= 0 + x/1! + 0x^2/2! - x^3/3!

= x - x^3/6

Therefore, the Taylor polynomials T2(x) and T3(x) for f(x) = sin(x) centered at x = 0 are

T2(x) = x

T3(x) = x - x^3/6

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The curvature of the curve r(t) is [tex]\sqrt( 5 + 2t cos t + t^2 ) / (2 + t^2)^{(3/2)}[/tex], and the radius of curvature is [tex](2 + t^2)^{(3/2)} / \sqrt( 5 + 2t cos t + t^2 ).[/tex]

To find the curvature and radius of curvature of the curve r(t) = (cos t + sin t) i + (sin t - t cos t) j, t > 0, we need to compute the first and second derivatives of the curve:

r(t) = (cos t + sin t) i + (sin t - t cos t) j

r'(t) = (-sin t + cos t) i + (cos t + t sin t) j

r''(t) = (-cos t - sin t) i + (2cos t + t cos t - sin t) j

The magnitude of the vector r'(t) is:

[tex]| r'(t) | = \sqrt( (-sin t + cos t)^2 + (cos t + t sin t)^2 )= \sqrt( 2 + t^2 )[/tex]

The curvature k is given by:

[tex]k = | r''(t) | / | r'(t) |^3[/tex]

Substituting the expressions for r'(t) and r''(t), we get:

[tex]k = | r''(t) | / | r'(t) |^3[/tex]

[tex]= | (-cos t - sin t) i + (2cos t + t cos t - sin t) j | / (2 + t^2)^{(3/2)}[/tex]

[tex]= \sqrt( (cos t + sin t)^2 + (2cos t + t cos t - sin t)^2 ) / (2 + t^2)^{(3/2)}[/tex]

Simplifying this expression, we get:

[tex]k = \sqrt( 5 + 2t cos t + t^2 ) / (2 + t^2)^{(3/2)}[/tex]

To find the radius of curvature R, we use the formula:

R = 1 / k

Substituting the expression for k, we get:

[tex]R = (2 + t^2)^{(3/2)} / \sqrt( 5 + 2t cos t + t^2 )[/tex]

Therefore, the curvature of the curve r(t) is [tex]\sqrt( 5 + 2t cos t + t^2 ) / (2 + t^2)^{(3/2)}[/tex], and the radius of curvature is [tex](2 + t^2)^{(3/2)} / \sqrt( 5 + 2t cos t + t^2 ).[/tex]

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The total length is 102.986 units

The total area of the floor sheet is 1020 units.

The method for finding the total area will depend on the shape(s) involved. Here are some general formulas for common shapes:

Square: To find the area of a square, multiply the length of one side by itself. For example, if the side of a square is 4 units, the area would be 4 x 4 = 16 square units.

Rectangle: To find the area of a rectangle, multiply the length by the width. For example, if a rectangle has a length of 6 units and a width of 4 units, the area would be 6 x 4 = 24 square units.

Triangle: To find the area of a triangle, multiply the base by the height and divide by 2. For example, if a triangle has a base of 5 units and a height of 8 units, the area would be (5 x 8) / 2 = 20 square units.

Circle: To find the area of a circle, multiply pi (approximately 3.14) by the radius squared. For example, if a circle has a radius of 3 units, the area would be 3.14 x 3^2 = 28.26 square units.

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If a straight line joins the points A (2, 1) and B (8, 10) and the point C (6, y) lies on the line AB. Then the y-coordinate of C is 7.

A line is a geometric object in mathematics that extends infinity in both directions and is symbolised by a straight line that never ends.

Two points, referred to as endpoints, define it as being one-dimensional and lacking in both width and depth.

A line equation has the slope-intercept form y = mx + b, where m is the slope and b is the y-intercept.

A number of real-world situations, such as the motion of an object or the direction of a force, can be modelled and described using lines.

We can find the equation of line AB using the two given points A and B:

Slope of AB = (change in y) / (change in x) = (10 - 1) / (8 - 2) = 9/6 = 3/2

Using point-slope form with point A:

y - 1 = (3/2)(x - 2)

Simplifying, we get:

y - 1 = (3/2)x - 3

y = (3/2)x - 2

Now we substitute x = 6 to find the y-coordinate of C:

y = (3/2)(6) - 2

y = 7

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This series is an arithmetic series, as the difference between consecutive terms is a constant, the series goes to infinity, meaning it has an infinite number of terms. Since the arithmetic series has an infinite number of terms, it is divergent. Therefore, the answer is DIVERGES.

To determine whether the series ∑n=1∞(4n+1)/(5−n) is convergent or divergent, we can use the ratio test:

lim┬(n→∞)⁡|((4(n+1)+1)/(5-(n+1)))/((4n+1)/(5-n))|

= lim┬(n→∞)⁡|(4n+5)(5-n)/(4n+1)(6-n)|

= 4/6 = 2/3

Since the limit is less than 1, the series is convergent by the ratio test.

To find the sum of the series, we can use the formula for the sum of a convergent geometric series:

S = a/(1-r)

where a is the first term and r is the common ratio. In this case, we have:

a = (4(1)+1)/(5-1) = 5/4

r = (4(2)+1)/(5-2) / (4(1)+1)/(5-1) = 13/6

Therefore, the sum of the series is:

S = (5/4) / (1-(13/6)) = 15/23
The given series is:

∑(4n + 15 - n) from n=1 to infinity.

First, simplify the series:

∑(3n + 15) from n=1 to infinity.

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The orthogonal decomposition of v with respect to w is [1, -1].

To find the orthogonal decomposition of v with respect to w, we need to find the orthogonal projection of v onto w, and the projection of v onto the orthogonal complement of w.

First, let's find the orthogonal projection of v onto w. The formula for the orthogonal projection of v onto w is:

[tex]\proj w(v) = ((v \cdot w)/||w||^2) * w[/tex]

where · represents the dot product and ||w|| represents the magnitude of w.

We have:

v = [1, -1]

w = [2, 2]

The dot product of v and w is:

v · w = 1*2 + (-1)*2 = 0

The magnitude of w is:

[tex]||w|| = \sqrt(2^2 + 2^2) = 2\sqrt(2)[/tex]

Therefore, the orthogonal projection of v onto w is:

[tex]\proj w(v) = ((v \cdot w)/||w||^2) * w = (0/(2\sqrt(2))^2) * [2, 2] = [0, 0][/tex]

Next, let's find the projection of v onto the orthogonal complement of w. The orthogonal complement of w is the set of all vectors that are orthogonal to w.

We can find a basis for the orthogonal complement of w by solving the equation:

w · x = 0

where x is a vector in the orthogonal complement. This equation represents the condition that x is orthogonal to w.

We have:

w · x = [2, 2] · [x1, x2] = 2x1 + 2x2 = 0

Solving this equation for x2, we get:

x2 = -x1

So any vector of the form [x1, -x1] is in the orthogonal complement of w. Let's choose [1, -1] as a basis vector for the orthogonal complement.

To find the projection of v onto [1, -1], we can use the formula:

[tex]\proj u(v) = ((v \cdot u)/||u||^2) * u[/tex]

where u is a unit vector in the direction of [1, -1]. We can normalize [1, -1] to obtain:

[tex]u = [1, -1]/||[1, -1]|| = [1/\sqrt(2), -1/\sqrt(2)][/tex]

The dot product of v and u is:

[tex]v \cdot u = 1*(1/\sqrt(2)) - 1*(-1/\sqrt(2)) = \sqrt(2)[/tex]

The magnitude of u is:

[tex]||u|| = \sqrt((1/\sqrt(2))^2 + (-1/\sqrt(2))^2) = 1[/tex]

Therefore, the projection of v onto [1, -1] is:

[tex]\proj u(v) = ((v \cdot u)/||u||^2) * u = (\sqrt(2)/1^2) * [1/\sqrt(2), -1/\sqrt(2)] = [1, -1][/tex]

Finally, the orthogonal decomposition of v with respect to w is:

v =[tex]\proj w(v) + \proj u(v)[/tex] = [0, 0] + [1, -1] = [1, -1]

Therefore, the orthogonal decomposition of v with respect to w is [1, -1].

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To test H0: (p1 - p2) = 0 against Ha: (p1 - p2) ≠ 0 with α = 0.07, follow these steps:

1. Calculate the sample proportions: p1_hat = 105/700 and p2_hat = 341/700.
2. Calculate the pooled proportion: p_pool = (105 + 341) / (700 + 700).
3. Calculate the standard error: SE = sqrt[(p_pool*(1-p_pool)/700) + (p_pool*(1-p_pool)/700)].
4. Calculate the test statistic (z): z = (p1_hat - p2_hat - 0) / SE.
5. Determine the rejection region: |z| > z_critical, where z_critical = 1.96 for α = 0.07 (two-tailed).
6. Compare the test statistic to the rejection region and make a conclusion.

The final conclusion: Based on the calculated test statistic and comparing it to the rejection region, we either reject or fail to reject the null hypothesis H0: (p1 - p2) = 0 at the 0.07 significance level, indicating that there is or isn't enough evidence to conclude that the population proportions are significantly different.

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-Q->P is proved from the contrapositive of 1.-P->Q.

What is Contrapostive ?

In logic, the contrapositive of a conditional statement of the form "if A, then B" is a statement of the form "if not B, then not A".

Here is a possible formal proof of the contrapositive of 1.-P->Q, which is -Q->P:

-P->Q Premise

Assume -Q Assume for sub proof

Assume -P Assume for subproof

From 2 and 1, we have P Modus tollens (MT)

From 4, we have P and -P Conjunction (CONJ)

From 3 and 5, we have # Contradiction (CONTR)

From 3-6, we have -Q-># Conditional proof (CP)

From 2 and 7, we have P Proof by contradiction (PC)

From 2-8, we have -Q->P Conditional proof (CP)

Therefore, -Q->P is proved from the contrapositive of 1.-P->Q.

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Answer:

Graph X

since it works ...............................................

The maximum profit Katie can earn is $85.

To find this, we need to use the vertex formula, which gives us the x-coordinate of the vertex of the parabola representing the profit function.

The formula is x = -b/2a, where a is the coefficient of the x-squared term (-1 in this case) and b is the coefficient of the x term (14 in this case). So, x = -14/(2*(-1)) = 7. Plugging this value into the profit function gives us P(7) = -7² + 14(7) + 57 = 85.

The profit function given is a quadratic function, which has a parabolic graph. The vertex of the parabola represents the maximum or minimum point of the function, depending on whether the leading coefficient is negative or positive. In this case, the leading coefficient is negative, so the vertex represents the maximum profit.

To find the x-coordinate of the vertex, we use the formula x = -b/2a, which gives us the value of x that makes the profit function equal to its maximum value. Once we have this value, we plug it back into the profit function to find the corresponding maximum profit. In this case, the maximum profit is $85.

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The maximum profit Katie can earn is $85.

To find this, we need to use the vertex formula, which gives us the x-coordinate of the vertex of the parabola representing the profit function.

The formula is x = -b/2a, where a is the coefficient of the x-squared term (-1 in this case) and b is the coefficient of the x term (14 in this case). So, x = -14/(2*(-1)) = 7. Plugging this value into the profit function gives us P(7) = -7² + 14(7) + 57 = 85.

The profit function given is a quadratic function, which has a parabolic graph. The vertex of the parabola represents the maximum or minimum point of the function, depending on whether the leading coefficient is negative or positive. In this case, the leading coefficient is negative, so the vertex represents the maximum profit.

To find the x-coordinate of the vertex, we use the formula x = -b/2a, which gives us the value of x that makes the profit function equal to its maximum value. Once we have this value, we plug it back into the profit function to find the corresponding maximum profit. In this case, the maximum profit is $85.

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Your answer is :-skew-symmetric.

We will show that if A is any n x n matrix, then (i) AAT is symmetric, and (ii) A - AT is skew-symmetric.

(i) To show that AAT is symmetric, we need to prove that (AAT)T = AAT. Here's the step-by-step explanation:

1. Compute the transpose of AAT, denoted as (AAT)T.
2. Using the reverse rule of transposes, we get (AAT)T = (AT)T * AT.
3. Since the transpose of a transpose is the original matrix, we have (AT)T = A.
4. Therefore, (AAT)T = A * AT = AAT, proving that AAT is symmetric.

(ii) To show that A - AT is skew-symmetric, we need to prove that (A - AT)T = -(A - AT). Here's the step-by-step explanation:

1. Compute the transpose of A - AT, denoted as (A - AT)T.
2. Using the properties of transposes, we get (A - AT)T = AT - A.
3. Now, we can observe that AT - A is the negation of A - AT, which is -(A - AT).
4. Therefore, (A - AT)T = -(A - AT), proving that A - AT is skew-symmetric.

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Answer: 18 pieces

Step-by-step explanation:

If she took 10 pieces and then each child received 2 pieces and she has four children then

10+2+2+2+2=

10+8=

18

The set corresponding to (a∪b)∩c is {3, 4, 5, 7, 9}.

To find the set corresponding to (a∪b)∩c, first perform the union of sets a and b, and then find the intersection with set c.

1. Union (a∪b):

a={x∈:x is a prime number}, b={4,7,9,11,13,14}.

Combine prime numbers and elements of set b:

{2, 3, 5, 7, 11, 13, 4, 7, 9, 11, 13, 14} and remove duplicates to get

{2, 3, 4, 5, 7, 9, 11, 13, 14}.

2. Intersection with c: (a∪b)∩c:

c={x∈:3≤x≤10}, meaning c={3, 4, 5, 6, 7, 8, 9, 10}.

Find the elements that are common between (a∪b) and c, so we get the set as {3, 4, 5, 7, 9}.

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Algebraically, the behaviour of the integral ∫(0 to ∞) (1/4x²) dx is that it diverges to infinity.

To determine the behaviour of the integral ∫(0 to ∞) (1/4x²) dx algebraically. Here's a step-by-step explanation:
1. First, rewrite the integral using proper notation:
  ∫(0 to ∞) (1/4x²) dx
2. Use the power rule for integration, which states that the integral of x^n is (x^(n+1))/(n+1), where n is a constant:
  ∫(1/4x²) dx = -1/(4(x))
3. Now, we will evaluate the indefinite integral from 0 to ∞:
  -1/(4(∞)) - (-1/(4(0)))
4. As x approaches ∞, the value of -1/(4x) approaches 0:
  0 - (-1/(4(0)))
5. As x approaches 0, -1/(4x) approaches infinity. However, since this is an improper integral, we need to consider a limit:
  lim(a->0) (-1/(4(a))) = ∞
So, algebraically, the behaviour of the integral ∫(0 to ∞) (1/4x²) dx is that it diverges to infinity.

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The area of the triangle is 3.535533, whose vertices are  (0,0,0), (1,3,−1), and (2,1,1) .

Explanation: -

To find the area of the triangle with vertices (0,0,0), (1,3,-1), and (2,1,1), you can use the formula for the area of a triangle in 3D space:

Area = 0.5 * ||AB x AC||

Here, AB and AC are the vectors representing the sides of the triangle, and "x" denotes the cross product.

STEP 1:- Find the vectors AB and AC:
AB = B - A = (1,3,-1) - (0,0,0)

     = (1,3,-1)
AC = C - A = (2,1,1) - (0,0,0)

     = (2,1,1)

STEP 2: - Compute the cross product AB x AC:
AB x AC = (3 * 1 - (-1) * 1, -(1 * 1 -(- 1 )* 2), 1 * 1 - 3 * 2)
AB x AC = (3 + 1, 3, 1 - 6)

              = (4, 2, -5)

STEP 3: - Compute the magnitude of the cross product ||AB x AC||:
||AB x AC|| = √(4² + (-3)² + (-5)²)

                  = √(16 + 9 + 25)

                  = √50

                  =7.071060

STEP 4: - Calculate the area of the triangle:
Area = 0.5 * ||AB x AC|| = 0.5 * (7.071060)

Therefore, the area of the triangle is 3.535533.

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Answer:  [DISCLAIMER]: All answers are rounded to the nearest hundredth of a decimal

f(1) = 5.20 ; f(2) = 14.70 ; f(3) = 27 ; f(4) = 41.57

Step-by-step explanation:

f(1) = (1 + [2×1])[tex]^{3/2}[/tex]

f(1) = (1 + 2)[tex]^{3/2}[/tex]

f(1) = (3)[tex]^{3/2}[/tex]

f(1) ≈ 5.20

f(2) = (2 + [2×2])[tex]^{3/2}[/tex]

f(2) = (2 + 4)[tex]^{3/2}[/tex]

f(2) = (6)[tex]^{3/2}[/tex]

f(2) ≈ 14.70

f(3) = (3 + [2×3])[tex]^{3/2}[/tex]

f(3) = (3 + 6)[tex]^{3/2}[/tex]

f(3) = (9)[tex]^{3/2}[/tex]

f(3) = 27

f(4) = (4 + [2×4])[tex]^{3/2}[/tex]

f(4) = (4 + 8)[tex]^{3/2}[/tex]

f(4) = (12)[tex]^{3/2}[/tex]

f(4) ≈ 41.57

The probability that all four switches work is 0.0033 or approximately 0.33%.

Given that each switch has a probability of .24 of working, we can use the following steps:

1. Understand that the probability of all four switches working is the product of their individual probabilities, since they work independently.
2. Multiply the probabilities: .24 (switch 1) × .24 (switch 2) × .24 (switch 3) × .24 (switch 4).

The probability that each switch works is 0.24. Since the switches work independently, we can multiply the probabilities to find the probability that all four switches work:

P(all four switches work) = 0.24 x 0.24 x 0.24 x 0.24

P(all four switches work) = 0.00327648

Rounding to four significant figures, we get:

P(all four switches work) ≈ 0.0033

Your answer: The probability that all four switches work is .24 × .24 × .24 × .24 = 0.0033.

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The integral of 6 |x − 3| dx from 0 to 6 is equal to 54, which represents the area under the curve.

First, let's split the integral into two parts based on the absolute value function:

1. When x < 3, |x - 3| = 3 - x.
2. When x ≥ 3, |x - 3| = x - 3.

Now, we can rewrite the integral as the sum of two separate integrals:

Integral from 0 to 3 of 6(3 - x) dx + Integral from 3 to 6 of 6(x - 3) dx

Next, we'll evaluate each integral:

1. Integral from 0 to 3 of 6(3 - x) dx:
  a. Find the antiderivative: 6(3x - (1/2)x^2) + C
  b. Evaluate at the bounds: [6(3(3) - (1/2)(3)^2) - 6(3(0) - (1/2)(0)^2)] = 27

2. Integral from 3 to 6 of 6(x - 3) dx:
  a. Find the antiderivative: 6((1/2)x^2 - 3x) + C
  b. Evaluate at the bounds: [6((1/2)(6)^2 - 3(6)) - 6((1/2)(3)^2 - 3(3))] = 27

Finally, add the two integrals together:

27 + 27 = 54

So, the integral of 6 |x − 3| dx from 0 to 6 is equal to 54, which represents the area

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