1. Find the value of X and Y:

2. Find the value of X:

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:

C 15

Step-by-step explanation:

9 =  12 3/4

20 3/4 = 15

The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).

BODMAS stands for B - Bracket, O - order of Power, D - Division, M - Multiplication, A - Addition, and S - Subtraction.

To Arrange the expressions below in order from least to greatest. place the least at the top and the greatest at the bottom

72 / 8 - 2 x (3 + 1) equals 1

(72 / 8) - 2 x 3 + 1 equals 4

72 / (8 - 2) x 3 + 1 equals 37

72 / (8 - 2) x (3 + 1) equals 48

Thus, The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).

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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]

[tex]Heyo![/tex]

SaddySokka is here to help!!

Let's do this step-by-step explanation!

[tex](15)(-5)+(15)(-3)[/tex]

[tex]=-75+(15)(-3)[/tex]

[tex]=-75+-45[/tex]

[tex]=-120[/tex]

Answer:

[tex]-120[/tex]

Hopefully, this helps you!!

Have a great day!!

SaddySokka~

Answer:

-120

Step-by-step explanation:

15×[-5]+15×[-3]

Use BODMAS

-75+-45

-120

Answer:

64 pi

Step-by-step explanation:

32 is diameter

diameter is circumference

2 circles

so 2*32=64

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)

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:

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:

W = 4 cm and  L = 9 cm

Step-by-step explanation:

I don't see a question, but will assume the problem wants the length(L) and width(W) of the described rectangle.

Let L and W stand for Length and Width.

Area of a rectangle is given by L*W

We are told that L*W = 36 cm^2

We are also told that L = 4W-7 ["length of a rectangle is 7 cm less than four times its width"]

Substituting the second into the first equation:

L*W = 36 cm^2

(4W-7)*W = 36 cm^2          [L = 4W-7]

4W^2-7W - 36 cm^2 = 0

(W-4)(4W+9) = 0

The roots are:  4 and -(9/4)

We'll use the positive value:  W = 4

Since L = 4W-7:

   L = 4(4)-7

   L = 16-7

  L = 9 cm

Answer:

Step-by-step explanation:

1.  x -> opposite side of 48°

   o → hypotenuse

   b → adjacent side of  48°

[tex]\sf Sin \ 48^\circ = \dfrac{opposite \ side }{hypotenuse}\\\\\\0.7431 = \dfrac{15}{o}\\\\\\0.74 * o = 15\\\\\\ o = \dfrac{15}{0.74}\\\\\\[/tex]

o = 20.27

[tex]\sf cos \ 48^\circ = \dfrac{adjacent \ side }{hypotenuse}\\\\\\0.67 =\dfrac{b}{o}\\\\\\0.67=\dfrac{b}{20.27}[/tex]

b = 0.67*20.27

b = 13.58

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2) i → opposite side of 25°

n → adjacent side of 25°

[tex]\sf Sin \ 25 =\dfrac{i}{t}\\\\\\0.42=\dfrac{i}{30}\\\\\\0.42*30=i[/tex]

i = 12.6

[tex]\sf Cos \ 30^\circ =\dfrac{n}{t}\\\\0.91=\dfrac{n}{30}\\\\\\0.91*30 = n[/tex]

n = 27.3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

3) a → opposite side of 70°

e → adjacent side of 70°

[tex]Sin \ 70^\circ =\dfrac{a}{l}\\\\0.94 =\dfrac{a}{25}\\\\0.94*25=a[/tex]

a = 23.5

[tex]\sf Cos \ 70^\circ =\dfrac{e}{l}\\\\0.34=\dfrac{e}{25}\\\\0.34*25=e[/tex]

e = 8.5

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

4)

[tex]\sf Sin \ 52^\circ = \dfrac{x}{75}\\\\0.79*75=x\\[/tex]

x = 59.25

[tex]\sf Cos \ 52^\circ = \dfrac{z}{75}\\\\0.62*75 =z[/tex]

z = 46.5

The inequality x -4 > 2 uses a greater than symbol

All numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2

The inequality is given as:

x -4 > 2

Add 4 to both sides of the inequality

x - 4 + 4 > 2 + 4

Evaluate the sum

x > 6

The above means that only numbers greater than 6 are in the solution of the inequality.

Since the options are not given, I will give a general solution that all numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2

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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: x is -2.5 or [tex]-2\frac{1}{2}[/tex]

Step-by-step explanation:

We need to solve for x

-3=4x+7

Step 1) Subtract 7 from both sides

-3-7=4x+7-7

-10=4x

Step 2) Divide both sides by 4 to isolate x

-10=4x

[tex]\frac{-10}{4} =\frac{4x}{4} \\-2.5=x[/tex]

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

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:

its 13 or B

Step-by-step explanation:

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:

The number of degrees the temperature drops°

Step-by-step explanation:

hope this helps

and hope this is the answer you was looking for

pls mark brainliest

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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Answer:

It's right

Step-by-step explanation:

(dk how to show prove but thank?

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!! :)

Wind is caused by differences in the atmospheric pressure. When a difference in atmospheric pressure exists, air moves from the higher to the lower pressure area, resulting in winds of various speeds. On a rotating planet, air will also be deflected by the Coriolis effect, except exactly on the equator.

Hope this helped!

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]