What is the Pythagorean triplet whose one member is 14

Answers

Answer 1

Answer:

14, 48 and 50

Step-by-step explanation:


Related Questions

What’s the answer for this ?

Answers

Step-by-step explanation:

1)-1, 0, 1... do like this

NEED HELP ASAP!!!! Will give brainiest

Answers

2%(random sentence bc I need 20 characters)

8. Daisies and tulips are planted in a
garden. There are 11 fewer tulips
planted than daisies.
a. Write an expression that represents
the number of tulips in terms of the
number of daisies. Define any
variables used.
b. If 18 daisies are planted, how many
tulips are planted?

Answers

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]

I need this for school, please help!!

Answers

C. Firstly, add up the amount of students and you should get 352. I did the easy route and divided 352 by 8 and got 44. From there, I added the amount of teachers on each graph and C is the only one with the amount of 44 teachers.

Arrange the expressions below in order from least to greatest. place the least at the top and greatest at the bottom. ( 72 ÷ 8 ) − 2 × 3 1 72 ÷ ( 8 − 2 ) × 3 1 72 ÷ ( 8 − 2 ) × ( 3 1 ) 72 ÷ 8 − 2 × ( 3 1 )

Answers

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).

What is BODMAS?

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).

Learn more about BODMAS;

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Wind is
• air moving from areas of high pressure to areas of low pressure.
• air moving from areas of low pressure to areas of high pressure.
air moving from areas of high temperature to areas of low temperature.

Answers

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!

Nada drives her car at an average speed of 60 miles per hour. She is planning a drive between 150 and 180 miles. Write and solve an inequality to model how many hours Nada will be driving

Answers

The inequality model that can be used to solve the number of hours it would take Nada to drive is 2.5 ≤ x ≤3.

What is the number of hours it would take Nada to drive?

Average speed is the total distance travelled per time.

Average speed = total distance / total time

Time = total distance / average speed.

Time if she drives 150 miles = 150 / 60 = 2.5 hours

Time if she drives 180 miles = 180 / 60 = hours

2.5 ≤ x ≤3.

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If -3=4x+7 what is x?

Answers

4x+7=-3
4x=-3-7
4x=-10
X=-10/4
X=-2.5

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]

Help Me write the inequality that represents the number line! Sorry the picture is bad quality!

Answers

Answer:

x ≤ -6

Step-by-step explanation:

The line is moving towards the left, which means that the numbers are getting smaller. X is going to be less than something.

The dot on -6 is colored in. This means that -6 is included in our inequality. [X could equal -6]

So, we could write the equation x ≤ -6

A plane leaves an airport traveling at 400 mph in the direction n 45° e. a wind is blowing at 40 mph in the direction n 45° w. what is the ground speed of the plane?

Answers

The ground speed of the plane is 402 mph

We have given,

A plane leaves an airport traveling at 400 mph in the direction n 45° e. a wind is blowing at 40 mph in the direction n 45° w.

Now, Draw a right triangle with 400 and 40 as the two legs.

This is a right angle because 45 degrees NE is perpendicular to 45 degrees NW.

So, the two side lengths are 400 and 40

Using the Pythagorean theorem,

What is the Pythagorean theorem?

(a^2 + b^2 = c^2)

[tex]400^2+40^2=c^2\\\\c^2=160000+1600\\c^2=161600\\c=401.99[/tex]

Therefore the answer is 401.995, or 402mph.

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∠A and \angle B∠B are vertical angles. If m\angle A=(7x-6)^{\circ}∠A=(7x−6)



and m\angle B=(8x-27)^{\circ}∠B=(8x−27)



, then find the measure of \angle B∠B

Answers

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]

7. What value of c will make x2 – 20x + c
a perfect square trinomial?

Answers

29x for this answer

The circumference of a circle is 3π cm. What is the area of the circle?

Question 3 options:

9πcm2

1.5πcm2

6πcm2

2.25πcm2

Answers

circumference of circle: 2πr

2πr = 3πr = 1.5 cm

area of circle: πr^2

πr^2π(1.5)^22.25π

Which pair of expressions has equivalent values?
1^13 and 1^15
6^1and 9^1
7^8and 8^7
9- and 4^3

Answers

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

A function is given.
f(x) = (3*0.04*x) + x

What is f (850)?
f(850) =

Answers

The function f(x) = (3 * 0.04 * x) + x is a linear function, and the value of f(850) is 952

How to evaluate the function?

The function is given as:

f(x) = (3 * 0.04 * x) + x

Substitute 850 for x

f(850) = (3 * 0.04 * 850) + 850

Evaluate the product

f(850) = 952

Hence, the value of f(850) is 952

Read more about linear functions at:

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A kitchen can be broken into 2 rectangles. One rectangle has a base of 7 feet and height of 5 feet. The second rectangle has a base of 2 feet and height of 2 feet. One package of tile will cover 3 square feet. How many packages of tile will she need? 8 13 15 39

Answers

Answer:

its 13 or B

Step-by-step explanation:

15 × [-5] +15 × [-3] = solution

Answers

[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

Which expression is equal to 0.75×0.09

Answers

The answer is C 75/100 * 9/10

help me please need help​

Answers

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 temperature is dropping at a rate of five degrees per hour.

Let d represent the number of degrees the temperature drops.
Let t represent the number of hours that pass.

Which is the dependent variable?

Answers

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

What are the solutions of the equation (x + 2)2 + 12(x + 2) – 14 = 0? Use u substitution and the quadratic formula to
solve.

-8+5√2
O x=-6252
O x=-4+5√2
x=-2 +5√2

Answers

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

Samir recorded the grade-level and instrument of everyone in the middle school School of Rock below. Seventh Grade Students Instrument # of Students Guitar 6 Bass 4 Drums 6 Keyboard 7 Eighth Grade Students Instrument # of Students Guitar 9 Bass 9 Drums 9 Keyboard 10 Based on these results, express the probability that a seventh grader chosen at random will play an instrument other than drums as a fraction in simplest form.

Answers

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.

What is a probability?

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

In this problem:

There is a total of 6 + 4 + 6 + 7 = 23 seventh graders.Of those, 23 - 6 = 17 play instruments that are not the drum.

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.

More can be learned about probabilities at https://brainly.com/question/14398287

Which value is not a solution of the inequality x-4 symbol 2

Answers

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

How to determine the value not in the solution?

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

Read more about inequality at:

https://brainly.com/question/11234618

What are range, index of qualitative variation (IQV), interquartile range (IQR), standard deviation, and variance

Answers

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:

A right triangle includes one algae that measures 14º. what is the measure of the third angle

A 14º C 90º

B 76º D 104º

Answers

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!! :)

Identify the surface area of the cylinder to the nearest tenth. Use 3.14 for π.

Answers

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

The length of a rectangle is 7 cm less than four times its width. The area of the rectangle is 36 square cm

Answers

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

Hello Calculus!

Find the value

[tex]\\ \rm\Rrightarrow {\displaystyle{\int\limits_3^5}}(e^{3x}+7cosx-3tan^3x)dx[/tex]

Note:-

Answer with proper explanation required and all steps to be mentioned .

Answers

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

DerivativesDerivative Notation

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]:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals

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

[Integral] Rewrite [Integration Rule - Addition/Subtraction]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + \int\limits^5_3 {7 \cos x} \, dx - \int\limits^5_3 {3 \tan^3 x} \, dx[/tex][Integrals] Rewrite [Integration Property - Multiplied Constant]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^3 x} \, dx[/tex][3rd Integral] Rewrite:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution and u-solve.

1st Integral

Set u:
[tex]\displaystyle u = 3x[/tex][u] Differentiate [Derivative Properties and Rules]:
[tex]\displaystyle du = 3 \, dx[/tex][Bounds] Swap:
[tex]\displaystyle \left \{ {{x = 5 \rightarrow u = 3(5) = 15} \atop {x = 3 \rightarrow u = 3(3) = 9}} \right.[/tex]

3rd Integral

Set v:
[tex]\displaystyle v = \sec x[/tex][v] Differentiate [Trigonometric Differentiation]:
[tex]\displaystyle dv = \sec x \tan x \, dx[/tex][dv] Rewrite:
[tex]\displaystyle dx = \frac{1}{\sec x \tan x} \, dv[/tex]

Step 4: Integrate Pt. 3

Let's focus on the 3rd integral first.

Apply Integration Method [U-Solve]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \int\limits^{x = 5}_{x = 3} {\frac{v^2 - 1}{v}} \, dv[/tex][Integral] Rewrite [Integration Property - Addition/Subtraction]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \int\limits^{x = 5}_{x = 3} {v} \, dv - \int\limits^{x = 5}_{x = 3} {\frac{1}{v}} \, dv \Bigg)[/tex][Integrals] Apply Integration Rules [Reverse Power Rule and Logarithmic Integration]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 5}_{x = 3} - \ln | v | \bigg| \limits^{x = 5}_{x = 3} \Bigg)[/tex][v] Back-Substitute:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 x}{2} \bigg| \limits^{5}_{3} - \ln | \sec x | \bigg| \limits^{5}_{3} \Bigg)[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 5 - \sec^2 3}{2}- \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \Bigg)[/tex]

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

[Integrals] Substitute in integrals:
[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]

∴ 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]

                                                         [ tan²x = sec²x - 1 ]

[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]

                                                  apply substitution ... u

[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]

substitute back u = sec(x)

[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]

PLEASE HELP!!!!!!!!!!!

Answers

Answer:

The answer is B. [tex]\frac{2x+1}{x^{2}-7} ,x\neq \sqrt{7}[/tex]

Step-by-step explanation:

[tex]f(x)=2x+1\\g(x)=x^{2} -7[/tex]

Focusing on the denominator makes it equal to = 0

[tex]x^{2} -7=0\\x^{2} =7\\x=+\sqrt{7},-\sqrt{7}[/tex]

If the denominator equals zero then the equation ends up undefined. A number over zero, zero can not go divided into a number.

A large multiplex movie house has many theaters. The largest theater has 41 rows. There are 19 seats in the first row. Each row has two seats more than the previous row. How many total seats are there in this theater?

Answers

Answer:

101

Step-by-step explanation:

I time 41x2 than add 19

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