What is the area of a square enclosure that uses 320 meters of fencing?

Answer:

To find the interquartile range (IQR), ​first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

Step-by-step explanation:

Answer:

its 13 or B

Step-by-step explanation:

Answer:

-8+5√2

Step-by-step explanation:

(x+2)^2+12(x+2)–14=0

(x+2)^2=(x+2)(x+2)=x^2+4+4x

12(x+2)=12x+24

x^2+4+4x+12x+24-14=0

x^2+4x+12x+4+24-14=0

x^2+16x+14=0

quadratic formula

x = {-b +- square root of (b^2 – 4ac)} ÷ {2a}

a= 1

b = 16

c = 14

x = {-16 +- square root of (16^2 – 4*1*14)} ÷ {2*1}

x = {-16 +- square root of (256 – 56)} ÷ {2*1}

x = ((-16 +- square root of (200)) ÷ (2)

x = ((-16 +- 10√2)) ÷ (2)

x= -8+-5√2

keeping in mind that vertical angles are always congruent.

[tex]\stackrel{\measuredangle A}{7x-6}~~ = ~~\stackrel{\measuredangle B}{8x-27}\implies -6=x-27\implies 21=x~\hfill \underset{\measuredangle B}{\stackrel{8(21)~~ - ~~27}{141}}[/tex]

Using correlation coefficients, it is found that the correct option is given by:

B) There is a high level of association between the two variables.

In this problem, it is said that the correlation between them is high, that is, strong, hence there is a high level of association, which means that option B is correct.

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Answer:

volume of a cylinder = π r^2 *h

108π=π*r^2*h

108=r^2*h

12=r^2

3.464=r

radius =3.464

Step-by-step explanation:

Brainliest would be appreciated

Ask for questions!

Answer:

The length of the cylinder's radius is 3

Step-by-step explanation:

A cylinders radius is found by the formula, r= √ v/πh

So, 108π/πh.

Because there is pi in both the numerator and denominator, we can cancel them out. This leaves the equation to be : r = √ 108/12.

r = √ 108/12

108/12 is 9.

r = √9

r = 3

Answer: B. 76

Step-by-step explanation: A right triangle is 180 degrees. It has an angle of 90 since it is a right triangle. 90 + 14 = 104. 180 - 104 = 76

Answer:

B. 76 degrees

Step-by-step explanation:

EVERY triangle's angles add up to 180 degrees. We already know that since it's a right triangle, one of the angles equals 90 degrees (that's a right angle) and they give us the second angle measurement, 14 degrees. If we add those two angle measures together and subtract them from 180, we should get the measure of the third angle as our answer.

14 + 90 = 104

180 - 104 = 76

Therefore, the third and final angle in this right triangle equals 76, so your answer is B. I hope this helps! Have a lovely day!! :)

Answer:

ohh just use this formula

this is for Area- A= π r^2

this is for Circumference- C= 2 π r

Step-by-step explanation:

The circumference of the circle of the radius of 13 inches is, 41 inches

Circumference is the distance around the perimeter of a circular object. It is defined as the length of the circle that is found by multiplying the diameter of the circle by π (pi), which is approximately equal to 3.14.

The formula for the circumference of a circle is given by: C = 2πr,

Given that,

The diameter of the circle is 13 inches,

The radius of the circle can be calculated as follows:

r = d/2

  = 13 inches / 2

  = 6.5 inches

Using the formula for the circumference, we can calculate the circumference as follows:

C = 2πr

   = 2 × 3.14 × 6.5 inches

   = 40.76 inches

Rounding to the nearest whole number, the circumference of the circle is 41 inches.

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Answer:

C 15

Step-by-step explanation:

9 =  12 3/4

20 3/4 = 15

Answer:

It's right

Step-by-step explanation:

(dk how to show prove but thank?

Answer:

967.6

Step-by-step explanation:

967.6

967.12 in

Step-by-step explanation:

the formula for the area (surface area) of a cylinder is: A=2πrh+2πr2

to solve we need to determine the values

R= radius = half the diameter = 14/2 =7

D= diameter =14inches

H= height= 15 inches

plug in

A=2πrh+2πr2 = A=2π([tex]\frac{d}{2}[/tex])h+2π([tex]\frac{d}{2}[/tex])^2

= 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])(15) + 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])^2

= 2[tex]\pi[/tex](7)(15) + 2[tex]\pi[/tex](7)^2

=[tex]\pi[/tex]((2x7x15)+(2x7^2))

=[tex]\pi[/tex](210+98)

=[tex]308\pi[/tex]

=967.12 in

Answer:

A)

Step-by-step explanation:

Sum of all angles of triangle = 180

53 + 68  + x = 180

        121 + x  = 180

                 x = 180 - 121

                 x = 59°

Answer:

180 - 53 - 68 = 59

The third angle is 59°.

Step-by-step explanation:

Answer:

Step-by-step explanation:

The only factor you can use is H. This doesn't factor into anything nice (like the common factor would be a decimal which is usually not allowed when doing this kind of question.

Answer: The question is the answer. 15x + 4 or H

The total length of the fencing is the perimeter of the rectangular yard which is P = 260 feet

What is the Perimeter of a Rectangle?

The perimeter P of a rectangle is given by the formula, P=2 ( L + W) , where L is the length and W is the width of the rectangle.

Perimeter P = 2 ( Length + Width )

Given data ,

Let the perimeter of the rectangular yard be = P

Now , the equation will be

The width of the rectangular yard W = 55 feet

The length of the rectangular yard L = 75 feet

And , the required fencing = Perimeter of rectangular yard

Perimeter of rectangular yard P = 2 ( L + W )

Substituting the values in the equation , we get

Perimeter of rectangular yard P = 2 ( 55 + 75 )

Perimeter of rectangular yard P = 2 ( 130 )

Perimeter of rectangular yard P = 260 feet

Therefore , the value of P is 260 feet

Hence , the perimeter of yard is 260 feet

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Answer:

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]

General Formulas and Concepts:
Calculus

Differentiation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]

Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]

Derivative Rule [Basic Power Rule]:

Integration

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Integration Method: U-Substitution + U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx[/tex]

Step 2: Integrate Pt. 1

Step 3: Integrate Pt. 2

Identify variables for u-substitution and u-solve.

1st Integral

3rd Integral

Step 4: Integrate Pt. 3

Let's focus on the 3rd integral first.

Step 5: Integrate Pt. 4

Focus on the other 2 integrals and solve using integration techniques listed above.

1st Integral:

[tex]\displaystyle\begin{aligned}\int\limits^5_3 {e^{3x}} \, dx & = \frac{1}{3} \int\limits^5_3 {3e^{3x}} \, dx \\& = \frac{1}{3} \int\limits^{15}_9 {e^{u}} \, du \\& = \frac{1}{3} e^u \bigg| \limits^{15}_9 \\& = \frac{1}{3} \bigg( e^{15} - e^9 \bigg)\end{aligned}[/tex]

2nd Integral:
[tex]\displaystyle\begin{aligned}7 \int\limits^5_3 {\cos x} \, dx & = 7 \sin x \bigg| \limits^5_3 \\& = 7 \bigg( \sin 5 - \sin 3 \bigg)\end{aligned}[/tex]

Step 6: Integrate Pt. 5

∴ we have evaluated the integral.

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Answer:

1086950.36760

Formula's used:

[tex]\rightarrow \sf \int sin(ax+b)=-\dfrac{1}{a} cos(ax+b)+c[/tex]

[tex]\rightarrow \sf \int cos(ax+b)=\dfrac{1}{a} sin(ax+b)+c[/tex]

[tex]\rightarrow \sf \int \dfrac{1}{ax+b} =\dfrac{1}{a} ln|ax+b|+c[/tex]

[tex]\rightarrow \sf \int e^{ax+b}=\dfrac{1}{a} e^{ax+b} + c[/tex]

[tex]\rightarrow \bold{ ln|a| - ln|b| = ln|\frac{a}{b} | }[/tex]

Explanation:

[tex]\dashrightarrow \sf \int \left(e^{3x}+7cos\left(x\right)-3tan^3\left(x\right)\right)[/tex]

                        apply sum rule: [tex]\bold{\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}[/tex]

[tex]\dashrightarrow \sf \int \:e^{3x}dx+\int \:7\cos \left(x\right)dx-\int \:3\tan ^3\left(x\right)dx[/tex]

                Integrate simple followings first, using formula's given above

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-\int 3tan^3x[/tex]

                        Breakdown the component

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int tan^2x(tanx)[/tex]

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int (sec^2x-1)(tanx)[/tex]

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

for integration of [tex]\bold{\int (sec^2x-1)(tanx)}[/tex]

[tex]\dashrightarrow \int \dfrac{-1+u^2}{u}[/tex]

[tex]\dashrightarrow \sf \int \:-\dfrac{1}{u}+udu[/tex]

[tex]\dashrightarrow \sf - \int \dfrac{1}{u}du+\int \:udu[/tex]

[tex]\dashrightarrow -\ln \left|u\right|+\dfrac{u^2}{2}[/tex]

[tex]\dashrightarrow \sf-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}[/tex]

================================================= insert back

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\left(-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}\right)[/tex]   outcome after integrating

Now apply the given limits

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{3(5)}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{3(3)}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                                                                   simplify

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{15}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{9}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                          and group the variables

[tex]\sf \hookrightarrow \dfrac{e^{15}-e^9}{3}-\dfrac{3}{2\cos ^2\left(5\right)}+\dfrac{3}{2\cos ^2\left(3\right)}+7\sin \left(5\right)-7\sin \left(3\right)+3\ln \left(\dfrac{1}{\cos \left(5\right)}\right)-3\ln \left(-\dfrac{1}{\cos \left(3\right)}\right)[/tex]

value:

[tex]\sf \hookrightarrow 1086950.36760[/tex]

Answer:

-2

-2

y=-2

Step-by-step explanation:

Answer:

-2

-2

y=-2

Step-by-step explanation:

Answer:

1^13 and 1^15

Step-by-step explanation:

1 raised to anything is still just 1

so, 1^13 = 1 and 1^15 =1

Answer:

a) [tex]d-11 = t[/tex]

b) 7

Step-by-step explanation:

Let [tex]d[/tex] = daisies

Let [tex]t[/tex] = tulips

a) 11 fewer tulips than daisies: [tex]d-11 = t[/tex]

b) Substitute 18 into [tex]d[/tex] and solve for [tex]t[/tex].

[tex]18-11= t[/tex]

[tex]7 = t[/tex]

Answer:

64 pi

Step-by-step explanation:

32 is diameter

diameter is circumference

2 circles

so 2*32=64

Using it's concept, it is found that there is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem:

Hence:

[tex]p = \frac{17}{23}[/tex]

There is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.

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