Area of the Region Between Curves-Please help.

Area Of The Region Between Curves-Please Help.

Answers

Answer 1

Answer: 32/3

Step-by-step explanation:

To find the area of the region between the curves, you first want to find the interval of both curves. You can do that by setting both curves equal to each other.

[tex]-\frac{1}{2}x^2+2=\frac{1}{2} x^2-2[/tex]                   [subtract both sides by 2]

[tex]-\frac{1}{2}x^2= \frac{1}{2} x^2-4[/tex]                         [subtract both sides by [tex]\frac{1}{2} x^2[/tex]]

[tex]-x^2=-4\\x=2,-2[/tex]        

Now that we know the interval, we set this into an integral and subtract the top curve by the bottom curve. *Note: I can't type in -2 into the interval, therefore only a "-" will be displayed.

[tex]\int\limits^2_- {|-\frac{1}{2}x^2+2-(\frac{1}{2}x^2-2) | } \, dx[/tex]                     [distribute -1]

[tex]\int\limits^2_-{|-\frac{1}{2}x^2+2-\frac{1}{2}x^2+2 |} \, dx[/tex]                       [combine like terms]

[tex]\int\limits^2_- {|-x^2+4|} \, dx[/tex]                                          [solve integral]

[tex]\frac{32}{3}[/tex]

The area between the curves is 32/3. This was also the reason why we had the absolute values because area can never be negative.


Related Questions

Erin is considering joining one of 2 clubs. Club A has no registration fee, but charges $105 per month. Club B charges members $80 per month plus a one-time registration fee of $375. For how many months is club A the cheaper option? Use system of equations

Answers

Answer:

4 months

Step-by-step explanation:

Club A

registration fee: $0

monthly fee: $105

After every month, the total cost increases by $105.

month 0: $0

month 1: $105

month 2: $210

month 3: $315

month 4: $420

month 5: $525

month 6: $630

Club B

registration fee: $375

monthly fee: $80

Notice how Club B's total reaches Club A's total after 2 months.

month 0: $375

month 1: $455

month 2: $535

month 3: $615

month 4: $695

Answer I need help !!!!!!!!!!!

Answers

Answer:

Pay for the day = $ 123.25

Step-by-step explanation:

From the question given:

Monday morning:

Time in: 8:15

Time out: 12:15 pm

Monday afternoon:

Time in: 13:00

Time out: 17:30

Pay = $ 14.5 /hr

Next, we shall determine the number of hours of work in the morning. This is illustrated below:

Time in: 8:15

Time out: 12:15 pm

Difference in time = 12:15 – 8:15 = 4 hrs

Next, we shall determine the pay for the work done in the morning. This can be obtained as follow:

Pay = $ 14.5 /hr

Pay for work done in the morning

= 4 × 14.5 = $ 58

Next, we shall determine the number of hours of work in the afternoon. This is illustrated below:

Time in: 13:00

Time out: 17:30 pm

Difference in time = 17:30 – 13:00 = 4 hrs 30 minutes

Next, we shall convert 4 hrs 30 minutes to hours. This is illustrated below:

60 minutes = 1 hr

30 minutes = 30/60 = 0.5 hrs.

Therefore,

4 hrs 30 minutes = 4 + 0.5 = 4.5 hrs

Next, we shall determine the pay for the work done in the afternoon. This can be obtained as follow:

Pay = $ 14.5 /hr

Pay for work done in the afternoon

= 4.5 × 14.5 = $ 65.25

Finally, we shall determine the pay for the day as follow:

Pay for work done in the morning

= $ 58

Pay for work done in the afternoon

= $ 65.25

Pay for the day = pay for morning + pay for afternoon

Pay for the day = $ 58 + $ 65.25

Pay for the day = $ 123.25

Therefore, the pay for the day is

$ 123.25

I really help worth these question.

Answers

Answer:

[tex] \frac{7}{3} [/tex]

Step-by-step explanation:

Given that,

p = -6,

q = 6

r = -19

Plug in the above values to evaluate the expression, [tex] \frac{\frac{q}{2} - \frac{r}{3}}{\frac{3p}{6} + \frac{q}{6}} [/tex]

[tex] \frac{\frac{6}{2} - \frac{(-19)}{3}}{\frac{3(-6)}{6} + \frac{6}{6}} [/tex]

[tex] \frac{\frac{3}{1} - \frac{(-19)}{3}}{\frac{-3}{1} + \frac{1}{1}} [/tex]

[tex] \frac{\frac{9 -(-19)}{3}}{3 + 1} [/tex]

[tex] \frac{\frac{28}{3}}{4} [/tex]

[tex] \frac{28}{3}*\frac{1}{4} [/tex]

[tex] \frac{28*1}{3*4} [/tex]

[tex] \frac{7*1}{3*1} [/tex]

[tex] \frac{7}{3} [/tex]

PLEASE ANSWER I REALLY NEED AND PEOPLE DONT ANSWER :(


1: identify the terms, coefficients, and constants in the expression 14x + 19.

Answers

Answer:

term=14x+19

cofficient=14+19

constant=19

Answer:

Terms :

14x , 19

Coefficients:

14

Constants:

19

Hope it helps.

algebra 2
50 POINTS
HELP

Answers

Answer:

{-3, 2}U{2, 5}

Step-by-step explanation:

For an equation to be negative, it would need to be in a negative range (below the x-axis or the coordinates are negative y-values). Therefore, we can examine this question and see that the graph is negative when the function crosses the x-axis at -3 and it remains negative until you reach 2 on the x-axis.

Therefore, the first set of negative values is (-3, 2).

Secondly, applying the same logic as before, the function decreases at 2 and then touches the x-axis again at 5. Therefore, the second negative value would be (2, 5).

The negative values are {-3, 2}U{2, 5}.

Answer:

{-3, 2}U{2, 5}

Step-by-step explanation:

If a wind turbine makes 64 full revolutions every 1 minute, what is its angular speed?

Answers

Answer:

this is wind turbine angular speed

Step-by-step explanation:

given data

angular speed ω = 64 rpm

time = 1 min = 60 seconds

                                                                           

solution

we know that angular speed ω is expess as

ω = [tex]\frac{2\pi }{T}[/tex]    .........................1

ω = 64 × [tex]\frac{2\pi }{T}[/tex]

ω  = 6.70 rad/s

so this is wind turbine angular speed

Robert has available 400 yards of fencing and wishes to enclose a rectangular area. Express the areaAof the rectangle as a function of the widthwof the rectangle. For what value ofwis the arealargest? What is the maximum area?

Answers

Answer:

A) A = 200w - w²

B) w = 100 yards

C) Max Area = 10000 sq.yards

Step-by-step explanation:

We are told that Robert has available 400 yards of fencing.

A) we want to find the expression of the area in terms of the width "w".

Since width is "w", and perimeter is 400,if we assume that length is l, then we have;

2(l + w) = 400

Divide both sides by 2 gives;

l + w = 200

l = 200 - w

Thus, Area of rectangle can be written as;

A = w(200 - w)

A = 200w - w²

B) To find the value of w for which the area is largest, we will differentiate the expression for the area and equate to zero.

Thus;

dA/dw = 200 - 2w

Equating to zero;

200 - 2w = 0

2w = 200

w = 200/2

w = 100 yards

C) Maximum area will occur at w = 100.

Thus;

A_max = 200(100) - 100(100)

A_max = 10000 sq.yards

The equation a = 640 s gives the relationship between s square miles and a acres. Pam owns 4.5 square miles of farmland. How many acres does she own? a. 2,880 acres b. 288 acres c. 0.7 acres d. 7.03 acres

Answers

Answer:

A. 2880 acres

Step-by-step explanation:

Formula: a = 640 s

Given information: s = 4.5

--> a = 640 x 4.5 = 2880 (acres)

which fraction is less then 7/10

Answers

Answer:

Well, there is a lot of answers to that. Check explanation, please!

Step-by-step explanation:

For example, some basic fractions that are less than 1/7 are:

1/8 (Compare their size!)2/18Or 1/154

You can also set up a number line, and compare the fractions.

Hopefully, this answer helps! :D

What will be the remainder when 6x ^5+ 4x^4 -27x^
3
- 7x² + 27x + 3/2 is divided by (2x^2 - 3)
^2

Answers

Answer:

Remainder = (3145/8)x - 408

Step-by-step explanation:

We want to find the remainder when 6x^(5) + 4x⁴ - 27x³ - 7x² + 27x + 3/2 is divided by (2x² - 3)²

Let's expand (2x² - 3)² to give ;

(2x - 3)(2x - 3) = 4x² - 6x - 6x + 9 = 4x² - 12x + 9

So,we can divide now;

______________________

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

First of all, we'll divide the term with the highest power inside the long division symbol by the term with the highest power outside the division symbol. This will give;

3/2x³

______________________

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

We now subtract the new multiplied term beneath the original one from the original one to get;

3/2x³

______________________

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³+27x +3/2

We'll now divide the term in new polynomial gotten with the highest power by the term with the highest power outside the division symbol. This gives;

(3/2)x³ + (11/2)x²

______________________

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

We now subtract the new multiplied term beneath the immediate one from the immediate one to get;

(3/2)x³ + (11/2)x²

______________________

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³-7x²+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

(159/2)x³-(113/2)x²+27x+(3/2)

We'll now divide the term in new polynomial gotten with the highest power by the term with the highest power outside the division symbol. This gives;

(3/2)x³ + (11/2)x² + (159/8)x

______________________

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³-7x²+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

(159/2)x³-(113/2)x²+27x+(3/2)

(159/2)x³-(477/2)x²+(1431/8)x

We now subtract the new multiplied term beneath the immediate one from the immediate one to get;

(3/2)x³ + (11/2)x² + (159/8)x

______________________

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³-7x²+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

(159/2)x³-(113/2)x²+27x+(3/2)

(159/2)x³-(477/2)x²+(1431/8)x

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

182x²-(1215/8)x + (3/2)

We'll now divide the term in new polynomial gotten with the highest power by the term with the highest power outside the division symbol. This gives;

(3/2)x³+(11/2)x²+(159/8)x+(91/2)

______________________

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³-7x²+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

(159/2)x³-(113/2)x²+27x+(3/2)

(159/2)x³-(477/2)x²+(1431/8)x

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

182x²-(1215/8)x + (3/2)

182x²-545x + 819/2

We now subtract the new multiplied term beneath the immediate one from the immediate one to get;

182x² - (1215/8)x + (3/2) - 182x² + 545x - 819/2 = (3145/8)x - 408

Remainder = (3145/8)x - 408

Find the slope and y-intercept (if possible) of the equation of the line. 14x − 6y = 66

Answers

Answer:

The slope would be 7/3

The y-intercept would be -11

Step-by-step explanation:

To find the answer rearrange your equation.

First subtract 14x from both sides, getting -6y= -14x + 66.

Then divide both sides by -6, getting y= 7/3x -11

This is now written in y=mx+b form. Your m is your slope and b is the y intercept!

Round to the nearest cent.
6. $10.407

Answers

Answer:

the answer is 10.41. If you have a number 5 or more you round the nearest number on the left up 1 if it's 4 or less it stays the same it doesn't go up or down.

$10.41 is the correct answer

Find the velocity, acceleration, and speed of a particle with position function r(t)=⟨−8tsint,−8tcost,2t2⟩

Answers

Answer:

The answer is below

Step-by-step explanation:

Velocity is the rate of change of displacement. Velocity is the ratio of distance to time.

The velocity v(t) = [tex]\frac{d}{dt}r(t)[/tex]

Where r(t) is the position function

Given that:

r(t)=⟨−8tsint,−8tcost,2t²⟩

[tex]v(t)=\frac{d}{dt}r(t)= <-8tcost-8sint,8tsint-8cost,4t>[/tex]

Acceleration is the rate of change of velocity, it is the ratio of velocity to time. Acceleration a(t) is given as:

[tex]a(t)=\frac{d}{dt}v(t)= \frac{d}{dt} <-8tcost-8sint,8tsint-8cost,4t>\\=<8tsint-16cost,8tcost+16cost,4>\\\\a(t)=<8tsint-16cost,8tcost+16cost,4>[/tex]

Speed = |v(t)| = [tex]\sqrt{(-8tcost-8sint)^2+(8tsint-8cost)^2+(4t)^2}\\\\ =\sqrt{64t^2cos^2t+128tcostsint+64sin^2t+64t^2sin^2t-128tsintcost+64cos^2t+16t^2}\\ \\=\sqrt{64t^2cos^2t+64t^2sin^2t+64sin^2t+64cos^2t+16t^2}\\\\=\sqrt{64t^2(cos^2t+sin^2t)+64(sin^2t+cos^2t)+16t^2}\\\\=\sqrt{64t^2+64+16t^2}=\sqrt{80t^2+64}[/tex]

3. Michael goes on a Camel Safari in Africa.
He travels 5 km north, then 4 km east, and
then 5 km south.
Distance:
Displacement:

Answers

Answers:

Distance = 14 km

Displacement = 4 km east

==============================================

Explanation:

The total distance is simply the sum of the values given. So he traveled a total of 5+4+5 = 9+5 = 14 km overall.

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

The displacement is where a drawing may come in handy. Draw an xy axis. Place point A at the origin (0,0). This is where Michael starts his trip. Then move up to (0,5) to indicate he goes 5 km north. Call this point B.

Afterward, mark point C at (4,5) to show he has gone 4 km east. Finally, point D is at (4,0) because he went 5 km south

The displacement is only concerned with two points: The start and end point. Nothing else matters. We started at A(0,0) and ended at D(4,0). So we've gone 4 km east overall. The direction is important when it comes to displacement. Simply saying "4 km" isn't enough. So in a sense, we could take a shortcut from A directly to D, bypassing the other points B and C.

See the diagram below for a visual.

A fair die is rolled 72 times and the percentage of 6s is recorded. What is the probability that at most 10% of the rolls are 6s

Answers

Answer:

P ([tex]\hat p[/tex] ≤ 0.10)

Step-by-step explanation:

The probability in terms of statistics for this given problem can be written as follows.

Let consider X to the random variable that represents the number of 6's in 7 throws of a dice, then:

X [tex]\sim[/tex]Bin ( n = 72, p = 0.167)

E(X) = np

E(X) = 72× 0.167

E(X) = 12.024

E(X) [tex]\simeq[/tex]  12

p+q =1

q = 1 - p

q = 1 - 0.167

q = 0.833

V(X) = npq

V(X) = 72 × 0.167 × 0.833

V(X) = 10.02

V(X) [tex]\simeq[/tex] 10

∴ X [tex]\sim[/tex] N ([tex]\mu = 12, \sigma^2 =10[/tex])

⇒ [tex]\hat p = \dfrac{X}{n} \sim N ( p, \dfrac{pq}{n})[/tex]

where p = 0.167 and [tex]\dfrac{pq}{n}[/tex] = [tex]\dfrac{0.167 \times 0.833}{72}[/tex] = 0.00193

∴ P(at most 10% of rolls are 6's)

i.e

P ([tex]\hat p[/tex] ≤ 0.10)

is 27.14159 rational or irrational

Answers

Answer:

It´s rational

Step-by-step explanation:

27,14159 = 2714159/100000

Rational

411,500 science notation

Answers

Answer:

the answer is 4.115 x 10^5

Step-by-step explanation:

hope that helps

Tony ran 1/2 of a mile for 1/4 of an hour. How many miles per hour did he run? A)1.0 B)2.0 C)3.0 D)4.0 E)5.0

Answers

Answer:

2.0

Step-by-step explanation:

Answer:

B. 2.0 miles

Step-by-step explanation:

Tony ran 1/2 of a mile for 1/4 of an hour.

First, to make it easier, change each fraction into decimals:

1/2 = 0.5

1/4 = 0.25

It takes Tony 0.25 hours to run 0.5 miles.

You are solving for 1 hours worth. Multiply 4 to both terms:

0.25 hr x 4 = 1 hr

0.5 miles x 4 = 2.0 miles

B. 2.0 miles is your answer.

~

Use the Trapezoidal Rule and Simpson's Rule to approximate the value of the definite integral for the value of n. Round the answers to four decimal places and compare the results with the exact value definite integral.
∫9 4 √xdx,n=8.

Answers

Answer and Step-by-step explanation: The Trapezoidal and Simpson's Rules are method to approximate a definite integral.

Trapezoidal Rule evaluates the area under the curve (definition of integral) by dividing the total area into trapezoids.

The formula to calculate is given by:

[tex]\int\limits^a_b {f(x)} \, dx = \frac{b-a}{2n}[f(x_{0})+2f(x_{1})+2f(x_{2})+...+2f(x_{n-1})+f(x_{n})][/tex]

The definite integral will be:

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{9-4}{2.8}[2+2.\sqrt{5} +2.\sqrt{6} +2.\sqrt{7}+2.\sqrt{8}+3][/tex]

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{5}{16}[25.3193][/tex]

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = 7.9122[/tex]

Simpson's Rule divides the area under the curve into an even interval number of subintervals, each with equal width.

The formula to calculate is:

[tex]\int\limits^a_b {f(x)} \, dx = \frac{b-a}{3n}[f(x_{0})+4f(x_{1})+2f(x_{2})+...+2f(x_{n-2})+4f(x_{n-1})+f(x_{n})][/tex]

The definite integral will be:

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{9-4}{3.8}[2+4.\sqrt{5} +2.\sqrt{6} +4.\sqrt{7} +4\sqrt{8} +3][/tex]

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{5}{24}[40.7398][/tex]

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = 8.4875[/tex]

Calculating the definite integral by using the Fundamental Theorem of Calculus:

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \int\limits^9_4 {x^{\frac{1}{2} }} \, dx[/tex]

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{2.\sqrt[]{x^{3}} }{3}[/tex]

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{2.\sqrt[]{9^{3}} }{3}-\frac{2.\sqrt[]{4^{3}} }{3}[/tex]

[tex]\int\limits^9_4 {\sqrt{x} } \, dx = 12.6667[/tex]

Comparing results, note that Simpson's Rule is closer to the exact value, i.e., gives better approximation to the exactly value calculated by the fundamental theorem.

which equation represents a line that is parallel to y= -4x+3 and passes through the point (-3,2)

Answers

Answer:

The answer is

[tex]y = - 4x - 14[/tex]

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

To find the equation of the parallel line we must first find the slope of the original line

From the question

y = - 4x + 3

Comparing with the general equation above

Slope / m = - 4

Since the lines are parallel their slope are also the same

Slope of parallel line = (-3 , 2) and slope

- 4 is

[tex]y - 2 = - 4(x + 3) \\ y + 2 = - 4x - 12 \\ y = - 4x - 12 - 2[/tex]

We have the final answer as

[tex]y = - 4x - 14[/tex]

Hope this helps you

Answer:

y=-4x-14

Step-by-step explanation:

Item 1 Evaluate. 1 1 3 +2 2 5 3 3 8 2 11 15 3 11 15 3 11 30

Answers

Answer:

add the expressions 832

Find the equation of the line using the point-slope formula. Passes through the point (−5, 8) and is parallel to the graph y = 4/5x+1

Answers

Answer: y=4/5x+12

Step-by-step explanation:

Slope-intercept form: y=ax+b, where a is the slope and b is the y-intercept

When a line is parallel to another line, they will have the same slope. So the slope of the new line is also 4/5.

y=4/5x+b

To find the y-intercept, we simply plug in the point it passes through.

y=4/5x+b

8=4/5(-5)+b

8=-4+b

b=12

y=4/5x+12

Hope this helps!! :)

Please let me know if you have any question

Which x values is the graph below discontinuous

Answers

For any value of X where there’s a jumb is discontinues. -3,-1,1,3

Answer:

-3,-1,1,3,5

Step-by-step explanation:

evaluate the following -3 - (-8)

Answers

Answer:

The answer to the problem is 5.

Answer:

5

Step-by-step explanation:

[tex]-3-\left(-8\right)\\\\\mathrm{Apply\:rule}\:-\left(-a\right)=a\\=-3+8\\\\\mathrm{Add/Subtract\:the\:numbers:}\:\\-3+8\\=5[/tex]

The electrical resistance of a wire, R, varies directly as its length, L, and inversely as its cross sectional area, A. If the resistance of a wire is 0.08 ohm when the length is 100 ft and its cross-sectional area is 0.05 in^2 , what is the resistance of a wire whose length is 4000 ft with a cross-sectional area of 0.02in^2 ? a) Write the variation equation. (R=k(l/a) b) Determine the value of the quantity indicated.   ​

Answers

Answer:

see below

Step-by-step explanation:

a  R=k(L/A)

Substitute what we know into the equation

.08 = k (100/.05)

.08 = k 2000

Divide each side by 2000

.08/2000 = k

.00004 = k

R=.00004 (L/A)

b R=.00004 (L/A)

We know L = 4000 and A = .02

R=.00004 (4000/.02)

R = 8

n ΔABC, AB = 10 and BC = 5. Which expression is always true? A. 5 < AC < 10 B. AC = 5 C. 5 < AC < 15 D. AC = 10

Answers

Answer:

A. 5 < AC < 10

Step-by-step explanation:

If ∆ABC is a right angled triangle, we use the Pythagoras formula:

c² = a² + b²

Where c = longest side

When given sides AB, AC and BC, the formula becomes:

AB² = AC² + BC²

Where AB = Longest side

In the question,

AB = 10 and BC = 5.

10² = AC² + 5²

AC² = 10² - 5²

AC² = 100 - 25

AC² = 75

AC = √75

AC = 8.6602540378

Therefore, the expression that is always true = A. 5 < AC < 10

You have two recipes that together use a pound of butter. One recipe takes a 1/4 pound more than the other one. How much butter does each recipe use?

Answers

Answer:

First recipe needs = 3/8

Second recipe needs = 5/8

Step-by-step explanation:

Given:

Two recipes needs butter = 1 pound

One recipe needs 1/4 pound more

Find:

Each recipe needs butter

Computation:

Assume;

First recipe needs = x

Second recipe needs = x + (1/4)

Two recipes needs butter = First recipe needs + Second recipe needs

x + x + (1/4) = 1

2 x = 1 - (1/4)

2 x = 3/4

x = 3/8 pound

First recipe needs = 3/8

Second recipe needs = x + (1/4) = (3/8) + (1/4) = 5/8

Second recipe needs = 5/8

what is the percentage of the total variation in candy bar sales explained by the regression model? a. 78.39% b. 88.54% c. 100% d. 48.19%

Answers

Complete question :

A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:

City - - - - - - - Price ($) -- - Sales

River City - - 1.30 - - - - - - 100

Hudson - - - 1.60 - - - - - 90

Ellsworth - - - 1.80 - - - - - 90

Prescott - - - - 2.00 - - - - 40

Rock Elm - - 2.40 - - 38

Stillwater - - 2.90 - - 32

Answer:

78.39%

Step-by-step explanation:

Given the data :

Price (X) :

1.30

1.60

1.80

2.00

2.40

2.90

Sales (y) :

100

90

90

40

38

32

The percentage of the total variation in candy bar sales explained by the regression model can be obtained from the value of the Coefficient of determination(R^2) of the regression model. The Coefficient of determination is a value which ranges between 0 - 1 and gives the proportion of variation in the dependent variable which can be explained by the dependent variable.

R^2 value is obtained by getting the squared value of R(correlation Coefficient).

The R value obtained using the online R value calculator on the data is : - 0.8854

Hence, R^2 = (-0.8854)^2 = 0.7839

Expressing 0.7839 as a percentage ;

0.7839 × 100 = 78.39%

Number of times the individual changed jobs in the last 5 years is what kind of variable? A. This variable is a continuous numerical variable that is interval-scaled. B. This variable is a discrete numerical variable that is interval-scaled. C. This variable is a categorical variable that is ordinal-scaled. D. This variable is a discrete numerical variable that is ratio-scaled. E. This variable is a continuous numerical variable that is ratio-scaled. F. This variable is a categorical variable that is nominal-scaled.

Answers

Answer: D. This variable is a discrete numerical variable that is ratio-scaled.

Step-by-step explanation:

A Discrete variables are variables which are countable in a finite amount of time. For example, you can count the amount of money in your bank wallet, but same can’t be said for the money you have deposited in eveyones bank account as this is infinite.

So the number of times an individual changes job in a five years period is a perfect example of  a discrete numerical variable that is ratio scaled because it can be counted.

Find the area of the region that lies inside the first curve and outside the second curve.
r= 10cos( θ)
r= 5

Answers

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = [tex]\dfrac{1}{2}[/tex]

[tex]\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}[/tex]

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

[tex]A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta[/tex]

[tex]A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta[/tex]

[tex]A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}[/tex]

[tex]A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}[/tex]

[tex]A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}[/tex]

[tex]A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}[/tex]

[tex]A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}[/tex]

[tex]A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}[/tex]

[tex]A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}[/tex]

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

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