evaluate the function at the fiven values of the variables:
f(x)= 5x^2 + 5x *+ 3
a f (-3)
b f (-9)

Try this option:

if only one answer of five is correct, it means, the probability to choose it P=1/5=0.2.

If the student guesses randomly, it means, using the probability 0.2, he(she) can choose only 10*0.2=2 correct answers. To pass the test, the student must get 0.6*10=6 correct answers or more.

Answer:

Step-by-step explanation:

You know that fractions with the same value in numerator and denominator reduce to 1. This is true whether the value is a number, a variable expression, or some mix of those. That is ...

[tex]\dfrac{1760}{1760}=1\\\\\dfrac{3\,\text{mi}}{1\,\text{mi}}=\dfrac{3}{1}\cdot\dfrac{\text{mi}}{\text{mi}}=3[/tex]

This example should show you that you can treat units as if they were a variable.

So, the unit conversion process is the process of choosing combinations of numerator and denominator units so that all the units you don't want cancel, leaving only units you do want.

You're starting with a number than has "laps" in the numerator. To cancel that, you need to find a conversion factor with "lap" in the denominator. On the list you are given, the one that has that is ...

  [tex]\dfrac{1920\,\text{yd}}{1\,\text{lap}}[/tex]

Now, you have canceled laps, but you have yards. Also on your list of conversion factors is a ratio with yards in the denominator:

  [tex]\dfrac{3\,\text{ft}}{1\,\text{yd}}[/tex]

This will cancel the yards in the numerator from the previous result, but will give you feet in the numerator. You want miles, so you look for a conversion factor between feet and miles, with miles in the numerator. The one you find is ...

  [tex]\dfrac{1\,\text{mi}}{5280\,\text{ft}}[/tex]

These three conversion factors go into the blanks. When you form the product, you will get ...

  [tex]\dfrac{38.5\cdot 1920\cdot 3}{1\cdot 1\cdot 5280}\cdot\dfrac{\text{laps$\cdot$yd$\cdot$ft$\cdot$mi}}{\text{lap$\cdot$yd$\cdot$ft}}=42\,\text{mi}[/tex]

Answer:

Step-by-step explanation:

Answer:

Only choices B and C are correct.

Step-by-step explanation:

The square root parent function is [tex]f(x)= \sqrt{x}.[/tex]

The function [tex]f(x)[/tex] is not a linear function, therefore it's graph cannot be a straight line. This rules out choice A.

The function [tex]f(x)[/tex] is defined at [tex]x=0[/tex] because [tex]f(0)= \sqrt{0} =0[/tex]. This means that choice B is correct.

The function [tex]f(x)[/tex] is only real for values [tex]x\geq 0[/tex], because negative values of [tex]x[/tex] give complex values for [tex]f(x)[/tex]. This means that choice C is correct.

The function [tex]f(x)[/tex] can take only positive values which means it is confined to only the I quadrant, and is not defined in quadrants II, III, and IV. This rules out Choice D.

Therefore only choices B and C are correct.

Answer:

B & C

Step-by-step explanation:

B and C are the answers

Answer: The mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 is

[tex]113.9\text{ and }4.0[/tex] respectively.

Step-by-step explanation:

Given : The mean of sampled population : [tex]\mu = 113.9[/tex]

The standard deviation of sampled population : [tex]\sigma = 32.1[/tex]

We know that the mean and standard deviation of the sampling distribution of the sample mean for samples of size n is given by :_

[tex]\mu_s=\mu\\\\\sigma_s=\dfrac{\sigma}{\sqrt{n}}[/tex]

Now, the mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 will be :-

[tex]\mu_s=113.9\\\\\sigma_s=\dfrac{32.1}{\sqrt{64}}=4.0125\approx4.0[/tex]

The mean of the sample is 113.9, and the standard deviation of the sample is 4.0

The given parameters are:

[tex]\mu = 113.9[/tex] --- the population mean

[tex]\sigma = 32.1[/tex] --- the population standard deviation

[tex]n = 64[/tex] --- the sample size

The sample mean is calculated as:

[tex]\bar x = \mu[/tex]

So, we have:

[tex]\bar x = 113.9[/tex]

The sample standard deviation is calculated as:

[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]

This gives

[tex]\sigma_x = \frac{32.1}{\sqrt {64}}[/tex]

[tex]\sigma_x \approx 4.0[/tex]

Hence, the mean of the sample is 113.9, and the standard deviation of the sample is 4.0

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Answer:

The solution is:

[tex](\frac{137}{13}, -\frac{152}{13})[/tex]

Step-by-step explanation:

We have the following equations

[tex]3x - 2y =55[/tex]

[tex]-2x - 3y = 14[/tex]

To solve the system multiply by [tex]\frac{3}{2}[/tex] the second equation and add it to the first equation

[tex]-2*\frac{3}{2}x - 3\frac{3}{2}y = 14\frac{3}{2}[/tex]

[tex]-3x - \frac{9}{2}y = 21[/tex]

[tex]3x - 2y =55[/tex]

---------------------------------------

[tex]-\frac{13}{2}y=76[/tex]

[tex]y=-76*\frac{2}{13}[/tex]

[tex]y=-\frac{152}{13}[/tex]

Now substitute the value of y in any of the two equations and solve for x

[tex]-2x - 3(-\frac{152}{13}) = 14[/tex]

[tex]-2x +\frac{456}{13} = 14[/tex]

[tex]-2x= 14-\frac{456}{13}[/tex]

[tex]-2x=-\frac{274}{13}[/tex]

[tex]x=\frac{137}{13}[/tex]

The solution is:

[tex](\frac{137}{13}, -\frac{152}{13})[/tex]

Answer:

x = 411/39 and y = -152/13

Step-by-step explanation:

It is given that,

3x - 2y = 55    ----(1)

-2x - 3y = 14  ---(2)

To find the solution of given equations

eq(1)  * 2  ⇒

6x - 4y = 110  ---(3)

eq(2) * 3  ⇒

-6x - 9y = 42  ---(4)

eq(3) + eq(4)  ⇒

6x - 4y = 110  ---(3)

-6x - 9y = 42 ---(4)

  0 - 13y = 152

y = -152/13

Substitute the value of y in eq (1)

3x - 2y = 55    ----(1)

3x - 2*(-152/13) = 55

3x + 304/13 = 55

3x = 411/13

x = 411/39

Therefore x = 411/39 and y = -152/13

The percent of all teens are online, use social media, and have posted a photo of themselves are:

                     70% (approx)

It is given that:

95% of teenagers  use the Internet this means that the teens are online.

This means that 95%=0.95.

and 81% of online teens use some kind of social media.

i.e. 81%=0.81

and  Of online teens who use some kind of social media, 91% have posted a photo of themselves.

i.e. 91%=0.91

Now we are asked to find the percent of teens who are online, use social media, and have posted a photo of themselves

i.e. The percent is: (0.95×0.81×0.91)×100

                             =  70.0245%

which is approximately equal to 70%

Answer:

The percent of all teens are online, use social media, and have posted a photo of themselves 70% (approx)

Step-by-step explanation:

[tex]|\Omega|=52\\|A|=4\\\\P(A)=\dfrac{4}{52}=\dfrac{1}{13}[/tex]

The probability of getting the card of 7 in a deck will be  1 / 13.

Probability is defined as the ratio of the number of favourable outcomes to the total number of outcomes in other words the probability is the number that shows the happening of the event.

Probability = Number of favourable outcomes / Number of sample

Given that a card is selected at random from a standard deck of playing cards. The probability of getting a number 7 will be,

There are four 7 cards in the deck of 52 cards.

Number of favourable outcomes = 4

Number of sample = 52

The probability is,

P = 4 / 52

P = 1 / 13

Therefore, the probability of getting the card of 7 in a deck will be  1 / 13.

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The linear equation can be written as; 9x - y = -2

A linear equation is an equation that has the variable of the highest power of 1. The standard form of a linear equation is of the form Ax + B = 0.

We are given the linear equation  as;

y = 9x + 2

-9x + y = 2

A cannot be a negative:

-1(-9x + y = 2)

9x - y = -2

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Explanation:

Assuming the side of the boat is a straight line, it would constitute a transversal crossing the lines of the oars. When corresponding angles at a transversal are congruent, the lines being crossed are parallel. Since the oars are those lines, the oars are parallel.

[tex]5y^4 + 11y^2 + 2=\\5y^4+10y^2+y^2+2=\\5y^2(y^2+2)+1(y^2+2)=\\(5y^2+1)(y^2+2)[/tex]

For this case we must factor the following expression:

[tex]5y ^ 4 + 11y ^ 2 + 2[/tex]

We rewrite[tex]y^ 4[/tex] as[tex](y^ 2) ^ 2[/tex]:

[tex]5 (y ^ 2) ^ 2 + 11y ^ 2 + 2[/tex]

We make a change of variable:

[tex]u = y ^ 2[/tex]

So, we have:

[tex]5u ^ 2 + 11u + 2[/tex]

we rewrite the term of the medium as a sum of two terms whose product is 5 * 2 = 10 and whose sum is 11. Then:[tex]5u ^ 2 + (1 + 10) u + 2\\5u ^ 2 + u + 10u + 2[/tex]

We factor the highest common denominator of each group:

[tex]u (5u + 1) +2 (5u + 1)[/tex]

We factor [tex](5u + 1):[/tex]

[tex](5u + 1) (u + 2)[/tex]

Returning the change:

[tex](5y ^ 2 + 1) (y ^ 2 + 2)[/tex]

ANswer:

[tex](5y ^ 2 + 1) (y ^ 2 + 2)[/tex]

To fill the hemispherical tank with water through a hole in the base, the work required is 418.88 foot-pounds.

To calculate the work required to fill the hemispherical tank with water through a hole in the base, we can use the concept of work done against gravity.

The volume of the tank can be calculated using the formula for the volume of a hemisphere, which is (2/3)πr^3. In this case, the radius is given as 2 feet.

The weight of the water can be found by multiplying the volume by the weight-density of water, which is 62.4 pounds per cubic foot.

The work done is then the weight of the water multiplied by the height it is lifted, which is equal to the radius of the hemisphere.

So, the work required to fill the tank with water is (2/3)π(2^3)(62.4)(2) = 418.88 foot-pounds.

The work required to fill the hemispherical tank with water through a hole in the base is [tex]$\frac{1248\pi}{5}$[/tex] foot-pounds.

To solve this problem, we need to calculate the work done against gravity to fill the tank with water. The work done to lift a small layer of water at a height [tex]$h$[/tex] is given by the force required to lift it times the height  [tex]$h$[/tex]. The force is the weight of the water, which is the volume of the water times its weight-density.

Let's break down the tank into thin horizontal slices. Each slice has a thickness [tex]$dh$[/tex] and is at a height [tex]$h$[/tex] from the base. The radius of each slice is given by [tex]$r = \sqrt{2^2 - h^2}$[/tex], where [tex]$2$[/tex] feet is the radius of the tank.

The volume of each slice is the area of the slice times its thickness [tex]$dh$[/tex]. The area of the slice is [tex]$\pi r^2$[/tex], so the volume [tex]$dV$[/tex] is [tex]\pi (2^2 - h^2) dh[/tex].

The weight of the water in each slice is the volume times the weight-density of water, which is [tex]$62.4$[/tex] pounds per cubic foot. Therefore, the weight of the water in each slice is [tex]$62.4\pi (2^2 - h^2) dh$[/tex].

The work done to lift this slice to the height [tex]$h$[/tex] is the weight of the water times the height [tex]$h$[/tex], which is [tex]$62.4\pi h (2^2 - h^2) dh$[/tex].

To find the total work done to fill the tank, we integrate this expression from [tex]$h = 0$[/tex]  to[tex]$h = 2$[/tex] feet:

[tex]\[ W = \int_{0}^{2} 62.4\pi h (2^2 - h^2) dh \] \[ W = 62.4\pi \int_{0}^{2} h (4 - h^2) dh \] \[ W = 62.4\pi \int_{0}^{2} (4h - h^3) dh \] \[ W = 62.4\pi \left[ 2h^2 - \frac{h^4}{4} \right]_{0}^{2} \] \[ W = 62.4\pi \left[ 2(2)^2 - \frac{(2)^4}{4} \right] - 62.4\pi \left[ 2(0)^2 - \frac{(0)^4}{4} \right] \] \[ W = 62.4\pi \left[ 8 - 4 \right] \] \[ W = 62.4\pi \tims 4 \] \[ W = 249.6\pi \] \[ W = \frac{1248\pi}{5} \][/tex]

Therefore, the work required to fill the hemispherical tank with water through a hole in the base is [tex]$\frac{1248\pi}{5}$[/tex] foot-pounds.

To evaluate the function f(x) = 5x² + 7 at a given value, substitute that value into the function and simplify. The resulting answer is the function's value at that point. For the difference quotient f(x) - f(7)/(x - 7), the value is undefined if x = 7.

To evaluate the function f(x) = 5x² + 7 at a given value of the independent variable and simplify the results, we simply plug in the value of the independent variable into the function. For example, if we want to evaluate f at x = 3, we would calculate f(3) = 5(3)² + 7 = 5(9) + 7 = 45 + 7 = 52. Thus, the evaluated function at x = 3 is 52.

If we were to find the difference quotient f(x) - f(7)/(x - 7), we would need to plug in a value for x that is not 7, as the expression is undefined for x = 7. For any other value of x, we substitute x into the function, subtract f(7), and divide by (x - 7). If x = 7, the result is UNDEFINED.

Let's denote the events as follows:

- A: Module A fails

- B: Module B fails

Given probabilities:

[tex]\[ P(A) = 0.24 \][/tex]

[tex]\[ P(B) = 0.38 \][/tex]

We can use the addition rule of probability for mutually exclusive events to calculate the probabilities:

(a) Probability that only Module A fails:

[tex]\[ P(A \text{ only}) = P(A) = 0.24 \][/tex]

(b) Probability that only Module B fails:

[tex]\[ P(B \text{ only}) = P(B) = 0.38 \][/tex]

(c) Probability that both modules fail:

[tex]\[ P(\text{both}) = P(A \cap B) = P(A) \times P(B) = 0.24 \times 0.38 = 0.0912 \][/tex]

(d) Probability that neither module fails:

[tex]\[ P(\text{neither}) = 1 - (P(A) + P(B) - P(A \cap B)) \][/tex]

[tex]\[ = 1 - (0.24 + 0.38 - 0.0912) \][/tex]

[tex]\[ = 1 - 0.5288 \][/tex]

[tex]\[ = 0.4712 \][/tex]

(e) The probabilities should add up to 1, representing the total probability of all possible outcomes:

[tex]\[ P(A \text{ only}) + P(B \text{ only}) + P(\text{both}) + P(\text{neither}) = 0.24 + 0.38 + 0.0912 + 0.4712 = 1 \][/tex]

So, the probabilities add up to 1, as expected.

a. The probability that only Module A fails is 0.1488

b. The probability that only Module B fails is 0.2888

c. The probability that both modules fail is 0.0912

d. The probability that neither module fails is 0.4712

e. The probabilities add up to is 1

Let's denote the events as follows:

Event A: Module A fails

Event B: Module B fails

Given probabilities:

P(A) = 0.24

P(B) = 0.38

(a) To find the probability that only Module A fails:

[tex]\[ P(A \cap \neg B) = P(A) \times P(\neg B) \][/tex]

To calculate [tex]\( P(\neg B) \)[/tex], we subtract the probability of B from 1:

[tex]\[ P(\neg B) = 1 - P(B) = 1 - 0.38 = 0.62 \][/tex]

[tex]\[ P(A \cap \neg B) = 0.24 \times 0.62 = 0.1488 \][/tex]

(b) Similarly, to find the probability that only Module B fails:

[tex]\[ P(\neg A \cap B) = P(\neg A) \times P(B) \][/tex]

[tex]\[ P(\neg A) = 1 - P(A) = 1 - 0.24 = 0.76 \][/tex]

[tex]\[ P(\neg A \cap B) = 0.76 \times 0.38 = 0.2888 \][/tex]

(c) To find the probability that both modules fail:

[tex]\[ P(A \cap B) = P(A) \times P(B) = 0.24 \times 0.38 = 0.0912 \][/tex]

(d) To find the probability that neither module fails:

[tex]\[ P(\neg A \cap \neg B) = P(\neg A) \times P(\neg B) \][/tex]

[tex]\[ P(\neg A \cap \neg B) = 0.76 \times 0.62 = 0.4712 \][/tex]

(e) The probabilities add up to :

[tex]\[ P(A \cap \neg B) + P(\neg A \cap B) + P(A \cap B) + P(\neg A \cap \neg B) = 0.1488 + 0.2888 + 0.0912 + 0.4712 = 1 \][/tex]

Answer:

a) Quantitative discrete

b) Quantitative continuous

c) Qualitative

d) Quantitative continuous

e) Quantitative discrete

f) Qualitative

g) Qualitative

h) Quantitative continuous

i) Quantitative continuous

j) Quantitative discrete

Step-by-step explanation:

Okay, so an easy way to remember the difference between qualitative data and quantitative data:

Quantitative is measurable, it measures "QUANTITY."

Qualitative is categorical, meaning it has numeric value.

Now what about the difference between quantitative discrete and quantitative continuous.

Quantitative discrete there is NO IN BETWEEN. It counts things as whole numbers. The example would be your problem:

Number of tickets sold to a concert. Tickets are sold WHOLE, because you can't sell half a ticket.

Quantitative continuous has numbers in between, specifically, there could be decimals or fractions in between. Like in our problem:

Distance to the closest movie theater.

It can be 10.5m, 1.6 km. It does not necessarily have to be EXACTLY 1 Km away or 10 m away. So this is a continuous.

The type of data (Quantitative discrete, Quantitative continuous, or Qualitative) depends on whether a response involves counting, measuring, or categorizing. Examples were given to illustrate these different types of data, from counting tickets (Quantitative discrete) to measuring time (Quantitative continuous), to choosing a preferred product (Qualitative).

The type of data used to describe a response differs depending on the nature of the response. Let's identify them:

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Answer:

B.

Step-by-step explanation:

Plug in and see:

Let's try the first (x,y)=(5,7)

4(5)+5(7) <=20

20+35<=20 not true so not A

Let's try (1,1) for (x,y)

4(1)+5(1)<=20

4+5<=20 that is true so choice B looks good

For this case we must replace each of the points and verify if the inequality is met:

Point: (5,7)

[tex]4 (5) +5 (7) \leq20\\20 + 35 \leq20[/tex]

It is not fulfilled!

Point: (1,1)

[tex]4 (1) +5 (1) \leq20\\9 \leq20[/tex]

Is fulfilled!

Point: (0,5)

[tex]4 (0) +5 (5) \leq20\\0 + 25 \leq20[/tex]

It is not fulfilled!

Point: (8,0)

[tex]4 (8) +5 (0) \leq20\\32 \leq20[/tex]

It is not fulfilled!

Answer:

(1,1)

Answer:

12. $142,857.14

13. $77,704.82 . . . it is changed by the accumulated interest on the amount

Step-by-step explanation:

12. You want one year's interest on the endowment to be equal to $10,000. The principal (P) can be found by ...

  I = Prt

  10,000 = P·0.07·1

  10,000/0.07 = P ≈ 142,857.14

The endowment must be $142,857.14 to pay $10,000 in interest annually forever.

__

13. If the first award is in 10 years, we want the above amount to be the value of an account that has paid 7% interest compounded annually for 9 years. (The first award is 1 year after this amount is achieved.) Then we want the principal (P) to be ...

  142,857.14 = P·(1 +0.07)^9

  142,57.14/1.07^9 = P = 77,704.82

The endowment needs to be only $77,704.82 if the first award is made 10 years after the endowment date.

The initial amount required for endowment yielding $10,000 annually in perpetuity with a discount rate of 7% is $142,857.14. If the first payment is postponed to 10 years later, the present value of this amount is $72,975.85.

The question pertains to figuring out the initial amount needed to fund an endowment that would yield $10,000 annually, indefinitely, given a discount rate of 7%. This is a case for the use of perpetuity, a financial concept in which an infinite amount of identical cash flows occur continually. The formula for perpetuity is: Perpetuity = Cash flow / Discount rate. Thus, in this case, the amount to endow is $10,000 / 0.07 = $142,857.14.

For the second part of the question, if the first award will be given 10 years from today, the present value of the perpetuity needs to be factored in. The present value of a perpetuity starting at a future point is: Present Value of Perpetuity = Perpetuity / (1 + r)^n where 'r' is the discount rate and 'n' is the number of periods before the perpetuity starts. Here, it will be $142,857.14 / (1 + 7%)^10 = $72,975.85

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Divide by 2 for both of the numbers

32/2, 42/2

16/21

Answer is 16/21

Answer:

Divide both numbers by 2

32÷2 ,42÷2

Or, 16,21

answer is 16,21

[tex]\displaystyle\int\frac{4x^2-6}{(x+5)(x-2)(3x-1)}\,\mathrm dx[/tex]

You have a rational expression whose numerator's degree is smaller than the denominator's. This tells you you should consider a partial fraction decomposition. We want to rewrite the integrand in the form

[tex]\dfrac{4x^2-6}{(x+5)(x-2)(3x-1)}=\dfrac a{x+5}+\dfrac b{x-2}+\dfrac c{3x-1}[/tex]

[tex]\implies4x^2-6=a(x-2)(3x-1)+b(x+5)(3x-1)+c(x+5)(x-2)[/tex]

You can use the "cover-up" method here to easily solve for [tex]a,b,c[/tex]. It involves fixing a value of [tex]x[/tex] to make 2 of the 3 terms on the right side disappear and leaving a simple algebraic equation to solve for the remaining one.

So the integral we want to compute is the same as

[tex]\displaystyle\frac{47}{56}\int\frac{\mathrm dx}{x+5}+\frac{10}{35}\int\frac{\mathrm dx}{x-2}+\frac58\int\frac{\mathrm dx}{3x-1}[/tex]

and each integral here is trivial. We end up with

[tex]\displaystyle\int\frac{4x^2-6}{(x+5)(x-2)(3x-1)}\,\mathrm dx=\frac{47}{56}\ln|x+5|+\frac27\ln|x-2|+\frac5{24}\ln|3x-1|+C[/tex]

which can be condensed as

[tex]\ln\left|(x+5)^{47/56}(x-2)^{2/7}(3x-1)^{5/24}\right|+C[/tex]

Answer:

The limit of the function at x approaches to [tex]\frac{\pi}{2}[/tex] is [tex]-\infty[/tex].

Step-by-step explanation:

Consider the information:

[tex]\lim_{x \to \frac{\pi}{2}}\frac{cos(x)}{1-sin(x)}[/tex]

If we try to find the value at [tex]\frac{\pi}{2}[/tex] we will obtained a [tex]\frac{0}{0}[/tex] form. this means that L'Hôpital's rule applies.

To apply the rule, take the derivative of the numerator:

[tex]\frac{d}{dx}cos(x)=-sin(x)[/tex]

Now, take the derivative of the denominator:

[tex]\frac{d}{dx}1-sin(x)=-cos(x)[/tex]

Therefore,

[tex]\lim_{x \to \frac{\pi}{2}}\frac{-sin(x)}{-cos(x)}[/tex]

[tex]\lim_{x \to \frac{\pi}{2}}\frac{sin(x)}{cos(x)}[/tex]

[tex]\lim_{x \to \frac{\pi}{2}}tan(x)}[/tex]

Since, tangent function approaches -∞ as x approaches to [tex]\frac{\pi}{2}[/tex]

, therefore, the original expression does the same thing.

Hence, the limit of the function at x approaches to [tex]\frac{\pi}{2}[/tex] is [tex]-\infty[/tex].

The function cos(x)/(1-sin(x)) approaches an indeterminate form as x approaches π/2 from the right. By applying L'Hopital's Rule, we find that the limit is equivalent to the limit of tan(x) as x approaches π/2 from the right. However, tan(π/2) is undefined, so the limit does not exist.

In this question, we are asked to find the limit of the function cos(x)/(1-sin(x)) as x approaches π/2 from the right. We can't directly substitute x = π/2 because it makes the denominator zero, yielding an indeterminate form of '0/0'.

So, we use L'Hopital's Rule, which states that the limit of a ratio of two functions as x approaches a particular value is equal to the limit of their derivatives.

The derivative of cos(x) is -sin(x) and the derivative of (1-sin(x)) is -cos(x). Using L'Hopital's Rule, we can now re-evaluate our limit substituting these derivatives.
lim [x → (π/2)+] (-sin(x)/-cos(x)) = lim [x → (π/2)+] tan(x)

When substituting x = π/2 into tan(x), we realize that the tan(π/2) is undefined, so the answer is the limit does not exist.

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Answer:

6π - 9√3 unit^2

Step-by-step explanation:

Area of sector

= 1/6(π (6)^2

= 6π

Area of triangle = √3/4 (6)^2 = 9√3

Area of the shaded segment = Area of sector - Area of triangle

= 6π - 9√3 unit^2

to the risk of sounding redundant.

[tex]\bf \textit{area of a segment of a circle}\\\\ A=\cfrac{r^2}{2}\left(\cfrac{\pi \theta }{180}-sin(\theta ) \right)~~ \begin{cases} r=&radius\\ \theta =&angle~in\\ &degrees\\ \cline{1-2} r=&6\\ \theta =&60 \end{cases}\implies A=\cfrac{6^2}{2}\left(\cfrac{\pi 60}{180}-sin(60^o ) \right) \\\\\\ A=18\left( \cfrac{\pi }{3}-\cfrac{\sqrt{3}}{2} \right)\implies \boxed{A=6\pi -9\sqrt{3}}\implies \implies A\approx 3.26[/tex]

Answer:

side 1:  40; side 2: 20; side 3: 22

Step-by-step explanation:

It appears that sides 1 and 3 are based on side 2.  So side 2 is our "control" and we'll call it x.

If side 1 is 2 times side 2, then

side 1 = 2x.

If side 3 is 2 feet longer than side 2, then

side 3 = x + 2

We are given the perimeter value.  The perimeter of any shape is a measure around the outside of the shape.  Since a triangle has 3 sides, we add them together and set their sum equal to the perimeter we were given.  That looks like this:

x + 2x + x + 2 = 82

Combine like terms to get 4x + 2 = 82

Solve for x to get 4x = 80 and x = 20.  That means that side 2 measures 20 feet; side 1 measures twice that, so side 1 = 40 feet, and side 3 measures 2 more than side 2, so side 3 measures 22.  Let's add them all together and make sure the numbers work:

20 + 40 + 22 = 82

It works!  So we're all done!

Answer:

a^3b^4

Step-by-step explanation:

You have 3 a's, so that would be

[tex]a^{3}[/tex]

and you have 4 b's so that would be

[tex]b^4[/tex]

so putting it together gives you

[tex]a^3b^4[/tex]

Answer:

  2.50 per slice would be okay because most people would order 2 slices and that would have them give you an even 5.00. Net profit for this would be 464.00

Step-by-step explanation:

The specials advertised on Domino's website is 5.99 for each pizza, 8 slices.

The break even price is 1.34 per slice. This is important to know because a business never wants to take a loss.

To break even, the club would need to charge $2 per slice. However, for fundraising, they might charge $3 per slice, giving a net profit of $400. Breakeven analysis is crucial in business as it helps in decision making and financial planning.

Based on the information given, we see that the business club wants to purchase 50 pizzas from Domino's and resell them on campus. Assuming each pizza costs $16 (as per the example given for Authentic Chinese Pizza), the initial cost for the club would be $800 (50 pizzas x $16).

To calculate the break-even price per slice, we would need to divide the total cost by the total number of slices: $800 / (8 slices x 50 pizzas) = $2 per slice. However, since this is a fundraiser, the club might want to add in a margin to generate profit, so they might charge for example $3 per slice, resulting in a profit of $1 per slice, or $400 total.

The breakeven analysis is crucial in the development of new products because it helps businesses determine the minimum production and sales levels they must achieve to avoid losing money. This is useful for decision making and financial planning in any business operation.

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Answer:

  If x is a multiple of 4, then x/2 is even.

Step-by-step explanation:

An integer is even, if it is equal to 2n for some integer n. We want x/2 to be even, so ...

  x/2 = 2n

  x = 4n . . . . . multiply by 2

That is, x will be equal to 4n for some integer n. x is a multiple of 4.

To determine a sufficient condition for x/2 to be an even integer, we need to find values of x for which x/2 results in an even integer.

To find a sufficient condition for x/2 to be an even integer, we need to determine for which values of x the expression x/2 results in an even integer. Since an even integer is divisible by 2 without a remainder, a sufficient condition for x/2 to be an even integer is that x itself is divisible by 2 without a remainder. In other words, x should be an even integer.

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Answer:

[tex] log ( \frac { 5 x } { 4 y} ) \implies [/tex] [tex] log ( 5 ) + log ( x ) - log ( 4 ) + log ( y ) [/tex]

Step-by-step explanation:

We are given the following for which we are to expand the logarithm:

[tex] log ( \frac { 5 x } { 4 y} ) [/tex]

Expanding the log by applying the rules of expanding the logarithms by changing the division into subtraction:

[tex] log ( 5 x ) - log ( 4 y ) [/tex]

[tex] log ( 5 ) + log ( x ) - log ( 4 ) + log ( y ) [/tex]

[tex]\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log\left( \cfrac{5x}{4y} \right)\implies \log(5x)-\log(4y)\implies [\log(5)+\log(x)]-[\log(4)+\log(y)] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \log(5)+\log(x)-\log(4)-\log(y)~\hfill[/tex]

a. Parameterize [tex]\Gamma[/tex] by

[tex]\vec r(t)=(t,2t,t)[/tex]

with [tex]0\le t\le1[/tex]. The work done by [tex]\vec F[/tex] along [tex]\Gamma[/tex] is

[tex]\displaystyle\int_\Gamma\vec F\cdot\mathrm d\vec r=\int_0^1(1,-t,t)\cdot(1,2,1)\,\mathrm dt=\int_0^1(1-t)\,\mathrm dt=\boxed{\frac12}[/tex]

b. Break up [tex]\Gamma[/tex] into each component line segment, denoting them by [tex]\Gamma_1[/tex] and [tex]\Gamma_2[/tex], and parameterize each respectively by

both with [tex]0\le t\le1[/tex]. Then the work done by [tex]\vec F[/tex] along each component path is

[tex]\displaystyle\int_{\Gamma_1}\vec F\cdot\mathrm d\vec r_1=\int_0^1(1,t-1,t)\cdot(-1,1,1)\,\mathrm dt=\int_0^1(2t-2)\,\mathrm dt=-1[/tex]

[tex]\displaystyle\int_{\Gamma_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(1,-2t,1+t)\cdot(2,1,1)\,\mathrm dt=\int_0^1(3-t)\,\mathrm dt=\frac52[/tex]

giving a total work done of [tex]-1+\dfrac52=\boxed{\dfrac32}[/tex].

c. Parameterize [tex]\Gamma[/tex] by

[tex]\vec r(t)=(\cos t,\sin t,1)[/tex]

with [tex]0\le t\le2\pi[/tex]. Then the work done by [tex]\vec F[/tex] is

[tex]\displaystyle\int_\Gamma\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}(1,-\cos t,1)\cdot(-\sin t,\cos t,0)\,\mathrm dr=-\int_0^{2\pi}(\sin t+\cos^2t)\,\mathrm dt=\boxed{-\pi}[/tex]

Answer:

The sum  [tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n![/tex]  diverges ∵ of the Ratio Test.

General Formulas and Concepts:
Calculus

Limits

Series Convergence Tests

Step-by-step explanation:

Step 1: Define

Identify.

[tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n![/tex]

Step 2: Find Convergence

Since infinity is greater than 1, the Ratio Test defines the sum  [tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n![/tex]  to be divergent.

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Topic: AP Calculus BC (Calculus I + II)

Unit: Taylor Series

Answer:

The largest area is 125000 m²

The dimensions of the farmland are 250 m and 500 m

Step-by-step explanation:

* Lets pick the information from the problem

- The farmland is shaped a rectangle

- The farmland will be bounded on one side by a river

- The other three sides are bounded by a​ single-strand electric fence

- The length of wire is 1000 m

- Lets consider the width of the rectangle is x and the length is y

- The side which will be bounded by the river is y

∴ The perimeter of the farmland which will be bounded by the electric

  fence = x + x + y  = 2x + y

- We will use the wire to fence the farmland

∵ The length of the wire is 1000 m

∵ The perimeter of the farmland is equal to the length of the wire

∴ 2x + y = 1000

- Lets find y in term of x

∵ 2x + y = 1000 ⇒ subtract 2x from both sides

∴ y = 1000 - 2x

- Now lets find the area can enclose by the wire

∵ The area of the rectangle = length × width

∵ The width of the farmland is x and its length is y

∴ The area of the farmland (A) = x × y = xy ⇒ (2)

- Use equation (1) to substitute the value of y in equation (2)

∴ A = x (1000 - 2x) ⇒ simplify

∴ A = 1000 x - 2 x²

- To find the maximum area we will differentiate A with respect to x

  and equate the answer by zero to find the value of x which will make

  the enclosed area largest

* Lets revise the rule of differentiation

- If y = ax^n, then dy/dx = a(n) x^(n-1)

- If y = ax, then dy/dx = a

- If y = a, then dy/dx = 0 , where a is a constant

∵ A = 1000 x - 2 x² ⇒ (3)

- Differentiate A with respect to x using the rules above

∴ dA/dx = 1000 - 2(2) x^(2-1)

∴ dA/dx = 1000 - 4x

- Put dA/dx = 0 to find the value of x

∵ 1000 - 4x = 0 ⇒ add 4x to both sides

∴ 1000 = 4x ⇒ divide both sides by 4

∴ 250 = x

∴ The value of x is 250

- Lets substitute this value in equation 3 to find the largest area

∵ A = 1000 x - 2 x²

∴ A = 1000 (250) - 2(250)² = 125000 m²

* The largest area is 125000 m²

∵ The width of the farmland is x

∵ x = 250

∴ The width of the farmland = 250 m

- Substitute the value of x in the equation (1) to find y

∵ y = 1000 - 2x

∵ x = 250

∴ y = 1000 - 2(250) = 1000 - 500 = 500

∵ The length of the farm lend is y

∴ The length of the farm land = 500 m

* The dimensions of the farmland are 250 m and 500 m

To get the maximum area from a rectangular farmland bounded on one side by a river and on the other three sides by a single-strand electric fence with 1000 meters of wire, the dimensions should be 250m x 500m yielding a maximum area of 125,000 square meters.

This problem can be approached as a classic calculus maximization problem. The scenario mentioned in your question is about maximizing the area of a rectangle with a constant perimeter, this occurs when the rectangle is a square.

The three sides of your plot will consume the 1000 m of wire, if the length of each adjacent sides are x and y (where x is the length of the fence along the river, and y is the length of the other two fences), then

y + 2x = 1000

Substituting y from this equation into the area equation, A = xy(which results in A = x(1000-2x)), you can differentiate this equation with regards to x and set the derivative equal to zero to find the x-value that will give the maximum area. This happens when x = 250m,  y = 500m then the maximum area is 125,000 m².

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Answer: There is 21.25% of reduction in blood pressure.

Step-by-step explanation:

Since we have given that

Diastolic blood pressure reading previously = 80

Diastolic blood pressure reading now = 63

Decrement is given by

[tex]80-63\\\\=17[/tex]

So, percentage he reduced his blood pressure is given by

[tex]\dfrac{17}{80}\times 100\\\\=\dfrac{1700}{80}\\\\=21.25\%[/tex]

Hence, there is 21.25% of reduction in blood pressure.