Determine the intervals on which the function is increasing, decreasing, and constant.

A. Increasing x>0; Decreasing x<0

B. Decreasing on all real numbers

C. Increasing on all real numbers

D. Increasing x<0; Decreasing x>0

Answer:

-55 +11p

Step-by-step explanation:

-11(5-p)

Distribute

-11*5 -11*(-p)

-55 +11p

Answer:

25

Step-by-step explanation:

Assuming the equation is f(x) = x³ - 2

Plug in 3 for x

f(3) = 3³-2

= 27-2

=25

Answer:

25!

I hope it's helpful

Answer:

cylinder:

curved sa = 2πrh

=> rectangle, breath = h, width = perimeter of circle =  2πr

total sa = 2πrh+2πr^2

=> curved sa + area of 2 circles

cone:

curved sa = πrl

=> treated as a triangle, base = circumference of circle, height = slant height

total sa = πr^2+πrl

=> curved sa + area of base (circle)

Answer:

The profit is maximum when x = 40.

Step-by-step explanation:

Cost function, C = 40 x + 300

Revenue function, R = 80 x - 0.5 x^2

The profit function is

[tex]P = R - C\\\\P = 80 x - 0.5 x^2 - 40 x - 300\\\\P = - 0.5 x^2 + 40 x - 300\\\\\frac{dP}{dx} = - x + 40\\\\So, \frac{dP}{dx} = 0\\\\-x + 40 = 0 \\\\x = 40[/tex]

So, the profit is maximum when x = 40 .

Answer:

Step-by-step explanation:

Integration (4t√t+a)dt

[tex]\int \left ( 4 t\sqrt t +a \right )dt\\\\=\int\left ( 4 t^{(\frac{3}{2})} +a\right ) dt\\\\= 4\times 2\times \frac{t^{\frac{5}{2}}}{5} + a t\\\\= 8 \frac{t^{\frac{5}{2}}}{5} + a t[/tex]

Answer:

2.5 km left

The midpoint is half of 5, which is 2.5, so he'll still have 2.5 km left to complete

Answer:

2

Step-by-step explanation:

Answe YEAH BOIIIIII!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer:

54, 55, 56, 57, 58

Step-by-step explanation:

Answer:

56 problems

Step-by-step explanation:

Set up an equation.

[tex]\frac{3}{5}x<35<\frac{13}{20}x[/tex]

Why do we do this? We are told that she solved MORE than 60%, or [tex]\frac{3}{5}[/tex], and LESS than 65%, or [tex]\frac{13}{20}[/tex]. Therefore, if we set the TOTAL number of problems to x, we have an equation we can solve.

[tex]\frac{3}{5}x<35<\frac{13}{20}x\\[/tex]

Multiply all parts of the inequality by 20 to get rid of the denominators.

[tex]20*\frac{3}{5}x<20*35<20*\frac{13}{20}x\\ \\12x<700<13x[/tex]

Now we can solve TWO individual inequalities to isolate the x variable.

[tex]12x<700\\x<\frac{700}{12}\\x < 175/3\\x<58[/tex]

We can approximate 175/3 to about 58 (rounding down). We will sometimes round down when we have to deal with whole numbers.

The second inequality is as follows.

[tex]13x>700\\x>700/13\\x>53[/tex]

Therefore, we can combine the two inequalities.

[tex]53<x<58[/tex]

There were in between 53 and 58 questions. Since the number of questions must be a whole number, there can be 54, 55, 56, 57, OR 58. Why does 58 also work? When you plug 58 back into the original equation, you get that it STILL works. This is due to the fact that inaccuracies in computations allow you to round UP.

However, the last thing to keep in mind is that there are four sections with an equal number of questions. Meaning, the final answer has to be a multiple of four. The only multiple of 4 is 56; therefore, the final answer is 56.

9514 1404 393

Answer:

  2x +2y -8

Step-by-step explanation:

If the equal sides are 'a' and the third side is 'b', then the perimeter is ...

  P = a +a +b = 2a +b

The length of the third side is then ...

  b = P -2a . . . . . . subtract 2a from both sides

Substituting the given expressions, we find ...

  b = (10x +4y -18) -2(4x +y -5)

  b = 10x +4y -18 -8x -2y +10

  b = 2x +2y -8 . . . . the length of the third side

Answer:

H0 : μ ≥ 98.6

H1 : μ < 98.6

Step-by-step explanation:

The population mean temperature, μ = 98.6

The null hypothesis takes up the value of the population mean temperature as the initial truth ;

The alternative hypothesis on the other hand is aimed at using a sample size of 109 to establish if the mean temperature is less than the population mean temperature.

The hypothesis ;

Null hypothesis, H0 : μ ≥ 98.6

Alternative hypothesis ; H1 : μ < 98.6

Answer:

4.67 mph

Step-by-step explanation:

Speed with no wind = 14 mph

Let wind speed = w mph

Thus;

Speed with wind = 14 + w

Speed against the wind = 14 - w

Now, we are told that against the wind he bikes 10 miles.

Thus, from; time = distance/speed, we have;

Time = 10/(14 - w)

Also, we are told he biked 20 miles with the wind. Thus;

Time = 20/(14 + w)

We are told the times he used in both cases are the same.

Thus;

10/(14 - w) = 20/(14 + w)

Divide both sides by 10 to get;

1/(14 - w) = 2/(14 + w)

Cross multiply to get:

1(14 + w) = 2(14 - w)

14 + w = 28 - 2w

w + 2w = 28 - 14

3w = 14

w = 14/3

w = 4.67 mph

This is the solution to your question.

Answer:

Step-by-step explanation:

If D can paint the room in 6 hours, in 1 hour she gets [tex]\frac{1}{6}[/tex] of the room painted;

If C can paint the room in x hours, in 1 hour she gets [tex]\frac{1}{x}[/tex] of the room painted.

It takes 4 hours to paint it together. Setting up the classic work equation gives us

[tex]\frac{1}{6}+\frac{1}{x}=\frac{1}{4}[/tex] and we need to solve for x. Multiply everything through by the LCM which is 12x:

[tex]12x(\frac{1}{6}+\frac{1}{x}=\frac{1}{4})[/tex] making our equation simplify to

2x + 12 = 3x and solve for x:

12 = x

C can paint the room alone in 12 hours.

Answer:

Hence the probability that the mean weight of the sample of 55 cows would differ from the population mean by less than 12 lbs is 0.66545.

Step-by-step explanation:

Answer:

0.6826 = 68.26% probability that you have values in this interval.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

X~N(8, 1.5)

This means that [tex]\mu = 8, \sigma = 1.5[/tex]

What is the probability that you have values between (6.5, 9.5)?

This is the p-value of Z when X = 9.5 subtracted by the p-value of Z when X = 6.5. So

X = 9.5

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{9.5 - 8}{1.5}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a p-value of 0.8413.

X = 6.5

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{6.5 - 8}{1.5}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a p-value of 0.1587

0.8413 - 0.1587 = 0.6826

0.6826 = 68.26% probability that you have values in this interval.

Answer:

Perimeter = 18 + 9pi

Area = 81 - 20.25*pi

Step-by-step explanation:

Perimeter = 9 + 9 + 2(2 pi r)/2     The twos cancel out.

Perimeter = 18 + 9*pi

Area of the square = 9 * 9 = 81 cm^2

Area of the 2 semicircles = 2 * pi * r^2/2

r = d/2

d = 9

r = 9/2 = 4.5

Area of the 2 semicircles = 2 (pi * 4.5^2)/2

Area of the 2 semicircles = 20.25 pi

Area of the blue figure = 81 - 20.25 pi

Answer:

If you want the area of something with the sides 14cm and 7cm then it would be 98 cm.

Step-by-step explanation:

Area = length * width

Area = 14 cm * 7 cm

Area = 98 cm

Answer:

C

Step-by-step explanation:

In the graph given, we can expect the x axis to be horizontal and the y axis to be vertical. This means that the arm span represents y and the height represents x.

Therefore, if a girl on her team is 63 inches tall, we can say that y=x+2, and since height is x, y = 63 + 2 = 65

Given:

A clients rash measures 5 cm × 4 cm.

To find:

The measure in inches.

Solution:

We know that,

2.54 cm = 1 inch

1 cm [tex]=\dfrac{1}{2.54}[/tex] inch

1 cm [tex]\approx 0.3937[/tex] inch

Using this conversion, we get

5 cm [tex]=5\times 0.3937[/tex] inches

5 cm [tex]=1.9685[/tex] inches

4 cm [tex]=4\times 0.3937[/tex] inches

4 cm [tex]=1.5748[/tex] inches

Therefore, the measure in inches is 1.9685 inches × 1.5748 inches.

Answer:[tex]2\sqrt[4]{4}[/tex]

Step-by-step explanation:

[tex]64^{\frac{1}{4} }= \sqrt[4]{64}=\sqrt[4]{(2)(2)(2)(2)(4)}=2\sqrt[4]{4}[/tex]

Answer:

(2, -3) and r = 3.

Step-by-step explanation:

you can also plug this equation in desmos but I guess it's good to know how to do it also:

The equation of a circle is (x-h)^2 + (y-k)^2 = r^2

Now in order to make a perfect square on both sides, we need to do this:

First add 9 to both sides:

x^2 + 6x + 9 + y^2 -4y +4 = 9.

I purposely shifted it to show the perfect square created when you add 9 to both sides. Factor:

(x+3)^2 + y^2 - 4y + 4 = 9.

now the second bolded part is allso a perfect square. Factor:

(x+3)^2 + (y-2)^2 = 9

Based on the equation of a circle, the center must be at (2, -3) and the radius is the square root of 9 which is 3.

:)

Answer:

y = 3x + 5

Step-by-step explanation:

just by going through the options and plugging in values from the chart, you can guess and check

y = 3x + 5 is the only answer that works

5 = 0 + 5

11 = 6 + 5

23 = 18 + 5

etc.

hope this helps!

9514 1404 393

Answer:

  (320 +130π) cm²

Step-by-step explanation:

The perimeter of the base will be the sum of two radii and the arc length of a quarter circle:

  P = 2r +r(π/2) = r(2+π/2)

For a radius of 10 cm, the perimeter of the base is ...

  P = (10 cm)(2+π/2) = (20+5π) cm

The lateral area of the quarter-cylinder is the product of this perimeter and the height:

  LA = Ph = ((20 +5π) cm)(16 cm) = (320 +80π) cm²

__

The total base area is the area of a half-circle of radius 10 cm, so is ...

  BA = 1/2πr² = (1/2)π(10 cm)² = 50π cm²

The total surface area is the sum of the base area and the lateral area:

  SA = BA +LA = (50π +(320 +80π)) cm² = (320 +130π) cm²

Answer:

the total number of arrangements possible is 1,440 ways

Step-by-step explanation:

Given;

total number of kids = 5

total number of counselors, = 2

Since the counselors must sit together in any order, first treat them as a single option. This gives 6! possible arrangements for all the participants.

Also, If they can sit in any order, then the total possible arrangements = 2(6!)

                                            = 2( 6 x 5 x 4 x 3 x 2 x 1)

                                            = 1,440 ways

Therefore, the total number of arrangements possible is 1,440 ways

Seating arrangement is unique way in which people can sit. The number of seating arrangements possible in this case is 2520

If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in [tex]p \times q[/tex]  ways.

Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.

Thus, doing A then B is considered same as doing B then A

In such situations, we need to model the situation with the view point which can be evaluated mathematically.

For give case, we can see that there are in total 7 seats. And 5 kids are to sit on them, with 2 camp counselors.

So 7 people have to sit on 7 seats.

But it is given that two counselors must sit together.

Now firstly, two counselors can choose 2 seats out of 7 seats in [tex]^7C_2 = \dfrac{7 \times 6}{2 \times 1} = 21[/tex] ways.

Then , in the rest of the 5 seats, 5 kids can arrange themselves in 5! ways(using permutations).

We have:

[tex]n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1\\\\5! =5\times 4\times 3\times 2\times 1 = 120[/tex]

Since each of this 120 arrangement is for each of 21 ways of counselors sitting, thus, there are 120 times 21 ways of those 7 people to sit (using rule of product), or total [tex]120 \times 21 = 2520[/tex]

Thus,

The number of seating arrangements possible in this case is 2520

Learn more about seating arrangements here:

https://brainly.com/question/13605688

Answer:

The .jpeg file is the answer. Others are formulas that I use to solve.

Answer:

[tex]\displaystyle f(x) = \frac{x + 3}{x - 2}[/tex] has no relative extrema when the domain is [tex]\mathbb{R} \backslash \lbrace 2 \rbrace[/tex] (the set of all real numbers other than [tex]2[/tex].)

Step-by-step explanation:

Assume that the domain of [tex]\displaystyle f(x) = \frac{x + 3}{x - 2}[/tex] is [tex]\mathbb{R} \backslash \lbrace 2 \rbrace[/tex] (the set of all real numbers other than [tex]2[/tex].)

Let [tex]f^{\prime}(x)[/tex] and [tex]f^{\prime\prime}(x)[/tex] denote the first and second derivative of this function at [tex]x[/tex].

Since this domain is an open interval, [tex]x = a[/tex] is a relative extremum of this function if and only if [tex]f^{\prime}(a) = 0[/tex] and [tex]f^{\prime\prime}(a) \ne 0[/tex].

Hence, if it could be shown that [tex]f^{\prime}(x) \ne 0[/tex] for all [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace[/tex], one could conclude that it is impossible for [tex]\displaystyle f(x) = \frac{x + 3}{x - 2}[/tex] to have any relative extrema over this domain- regardless of the value of [tex]f^{\prime\prime}(x)[/tex].

[tex]\displaystyle f(x) = \frac{x + 3}{x - 2} = (x + 3) \, (x - 2)^{-1}[/tex].

Apply the product rule and the power rule to find [tex]f^{\prime}(x)[/tex].

[tex]\begin{aligned}f^{\prime}(x) &= \frac{d}{dx} \left[ (x + 3) \, (x - 2)^{-1}\right] \\ &= \left(\frac{d}{dx}\, [(x + 3)]\right)\, (x - 2)^{-1} \\ &\quad\quad (x + 3)\, \left(\frac{d}{dx}\, [(x - 2)^{-1}]\right) \\ &= (x - 2)^{-1} \\ &\quad\quad+ (x + 3) \, \left[(-1)\, (x - 2)^{-2}\, \left(\frac{d}{dx}\, [(x - 2)]\right) \right] \\ &= \frac{1}{x - 2} + \frac{-(x+ 3)}{(x - 2)^{2}} \\ &= \frac{(x - 2) - (x + 3)}{(x - 2)^{2}} = \frac{-5}{(x - 2)^{2}}\end{aligned}[/tex].

In other words, [tex]\displaystyle f^{\prime}(x) = \frac{-5}{(x - 2)^{2}}[/tex] for all [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace[/tex].

Since the numerator of this fraction is a non-zero constant, [tex]f^{\prime}(x) \ne 0[/tex] for all [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace[/tex]. (To be precise, [tex]f^{\prime}(x) < 0[/tex] for all [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace\![/tex].)

Hence, regardless of the value of [tex]f^{\prime\prime}(x)[/tex], the function [tex]f(x)[/tex] would have no relative extrema over the domain [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace[/tex].