Mathematics High School

A sequence can be generated by using a,n = a,n (-1) + 2, where a,1 = 6 and n is a whole number greater than 1.


What are the first 5 terms in the sequence?


A.) 6, 12, 24, 48, 96

B.) 2, 8, 14, 20, 26

C.) 2, 12, 72, 432, 2592

D.) 6, 8, 10, 12, 14

Answers

Answer 1
It’s b if it’s not ask your teacher

Related Questions

WHAT S THE LENGTH
OF THE MISSING SIDE?
3

Answers

Answer:

4

Step-by-step explanation:

using Pythagorean theorem

a^2+b^2=c^2

you have C so you can flip it around

c^2-b^2 will give you a^2

Answer:

the answer is 4

Step-by-step explanation:

b²=5²-3²

b²=16

b=4

Hello ! Here's a calculus question!

Find that value of

[tex]\\ \rm\Rrightarrow {\displaystyle{\int\limits_{\pi}^{2\pi}}}\dfrac{sin^6x+cos^6x}{sin^3xcos^3x}[/tex]


Note:-

Answer must include all steps properly .

Kindly don't waste time here if you don't know the answer .

All the best !​

Answers

Answer:

Undefined.

Formula's used:

[tex]\longrightarrow \bold{sec(x) = \dfrac{1}{cos(x)} }[/tex]

[tex]\longrightarrow \bold{cosec(x) = \dfrac{1}{sin(x)} }[/tex]

[tex]\longrightarrow \bold{cot(x) = \dfrac{1}{tan(x)} }[/tex]

[tex]\longrightarrow \bold{tan(x) = \dfrac{sinx}{cos(x)} }[/tex]

[tex]\longrightarrow \bold{sin^2x + cos^2 x = 1 }[/tex]

[tex]\longrightarrow \sf \bold{Sum \ Rule}:\quad \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]\longrightarrow \bold{ \int \:{sec^2(ax+b) = \dfrac{1}{a}tan(ax+b)+c }}[/tex]

[tex]\longrightarrow \bold{\int \:\dfrac{1}{ax+b} =\dfrac{1}{a} ln|ax+b|+c}[/tex]

Explanation:

[tex]\sf \Longrightarrow \int _{\pi }^{2\pi }\:\dfrac{sin^6(x)+cos^6(x)}{sin^3(x) \ * \ cos^3\left(x)}[/tex]

[tex]\sf \Longrightarrow \sf \int _{\pi }^{2\pi }\:\dfrac{sin^6\left(x\right)}{sin^3\left(x\right)\cdot \:cos^3\left(x\right)} +\dfrac{cos^6\left(x\right)}{sin^3\left(x\right)\cdot \:cos^3\left(x\right)}[/tex]

[tex]\Longrightarrow \sf \int _{\pi }^{2\pi }\:\dfrac{sin^3\left(x\right)}{\:cos^3\left(x\right)} +\dfrac{cos^3\left(x\right)}{sin^3\left(x\right)}[/tex]

                                                               

[tex]\Longrightarrow \sf \int _{\pi }^{2\pi }\ tan^3(x)+ cot^3(x)[/tex]

[tex]\sf \Longrightarrow \sf \int _{\pi }^{2\pi } \tan ^3(x)dx+\int _{\pi }^{2\pi } \cot ^3 (x)dx[/tex]

[tex]\Longrightarrow \sf \bold{ [ }-\ln |\sec \left(x\right) |+\dfrac{\sec ^2(x)}{2}-\dfrac{\cot ^2 (x)}{2}-\ln |\sin(x)| \bold{ ] }^{2\pi }_\pi[/tex]

apply limits

[tex]\sf \Longrightarrow \sf -\ln \left|\sec \left(2\pi \right)\right|+\dfrac{\sec ^2\left(2\pi\right)}{2}-\dfrac{\cot ^2\left(2\pi\right)}{2}-\ln \left|\sin \left(2\pi\right)\right|-(-\ln \left|\sec \left(\pi \right)\right|+\dfrac{\sec ^2\left(\pi\right)}{2}-\dfrac{\cot ^2\left(\pi\right)}{2}-\ln \left|\sin \left(\pi\right)\right|)[/tex]

simplify using trigonometric basic functions

[tex]\Longrightarrow \sf -\ln \left|\dfrac{1}{\cos \left(2\pi \right)}\right|+\dfrac{\left(\dfrac{1}{\cos \left(2\pi \right)}\right)^2}{2}-\dfrac{\cot ^2\left(2\pi \right)}{2}-\ln \left|\sin \left(2\pi \right)\right|- ( -\ln \left|\dfrac{1}{\cos \left(\pi \right)}\right|+\dfrac{\left(\frac{1}{\cos \left(\pi \right)}\right)^2}{2}-\dfrac{\cot ^2\left(\pi \right)}{2}-\ln \left|\sin \left(\pi \right)\right|)[/tex]

The value of cot(2π) is not defined. [ cot(2π) = ∞ ]

⇒  Undefined

Answer:

[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \boxed{\text{un} \text{de} \text{f}}[/tex]

General Formulas and Concepts:
Calculus

Differentiation

DerivativesDerivative Notation

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 Methods: U-Substitution + U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx[/tex]

Step 2: Integrate Pt. 1

[Integrand] Rewrite:
[tex]\displaystyle \frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x} = \frac{\sin^6 x}{\sin^3 x \cos^3 x} + \frac{\cos^6 x}{\sin^3 x \cos^3 x}[/tex][Integrand] Simplify:
[tex]\displaystyle \frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x} = \tan^3 x + \cot^3 x[/tex][Integrand] Rewrite:
[tex]\displaystyle \frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x} = \tan x (\sec^2 x - 1) + \cot x (\csc^2 x - 1)[/tex]


Step 3: Integrate Pt. 2

[Integral] Rewrite:
[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \int\limits^{2 \pi}_{\pi} {\tan x (\sec^2 x - 1) + \cot x (\csc^2 x - 1)} \, dx[/tex][Integral] Rewrite [Integration Property - Addition/Subtraction]:
[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \int\limits^{2 \pi}_{\pi} {\tan x (\sec^2 x - 1)} \, dx + \int\limits^{2 \pi}_{\pi} {\cot x (\csc^2 x - 1)} \, dx[/tex]

Step 4: Integrate Pt. 3

Identify variables for u-solve.

1st Integral

Set u:
[tex]\displaystyle u = \sec x[/tex][u] Apply Trigonometric Differentiation:
[tex]\displaystyle du = \sec x \tan x \, dx[/tex][du] Rewrite:
[tex]\displaystyle dx = \frac{1}{\sec x \tan x} \, du[/tex]

2nd Integral

Set v:
[tex]\displaystyle v = \csc x[/tex][v] Apply Trigonometric Differentiation:
[tex]\displaystyle dv = - \cot x \csc x \, dx[/tex][dv] Rewrite:
[tex]\displaystyle dx = \frac{-1}{\cot x \csc x} \, dv[/tex]

Step 5: Integrate Pt. 4

[Integrals] Apply Integration Method [U-Solve]:
[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \int\limits^{x = 2 \pi}_{x = \pi} {\frac{\tan x (\sec^2 x - 1)}{\sec x \tan x}} \, du + \int\limits^{x = 2 \pi}_{x = \pi} {\frac{- \cot x (\csc^2 x - 1)}{\cot x \csc x}} \, dv[/tex][Integrals] Simplify:
[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \int\limits^{x = 2 \pi}_{x = \pi} {\frac{u^2 - 1}{u}} \, du - \int\limits^{x = 2 \pi}_{x = \pi} {\frac{v^2 - 1}{v}} \, dv[/tex][Integrals] Apply Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \frac{u^2}{2} \bigg| \limits^{x = 2 \pi}_{x = \pi} - \int\limits^{x = 2 \pi}_{x = \pi} {\frac{1}{u}} \, du - \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 2 \pi}_{x = \pi} - \int\limits^{x = 2 \pi}_{x = \pi} {\frac{1}{v}} \, dv \Bigg)[/tex][Integrals] Apply Logarithmic Integration:
[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \frac{u^2}{2} \bigg| \limits^{x = 2 \pi}_{x = \pi} - \ln | u | \bigg| \limits^{x = 2 \pi}_{x = \pi} - \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 2 \pi}_{x = \pi} - \ln | v | \bigg| \limits^{x = 2 \pi}_{x = \pi} \Bigg)[/tex]Back-Substitute variables u and v:
[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \frac{\sec^2 x}{2} \bigg| \limits^{2 \pi}_{\pi} - \ln | \sec x | \bigg| \limits^{2 \pi}_{\pi} - \Bigg( \frac{\csc^2 x}{2} \bigg| \limits^{2 \pi}_{\pi} - \ln | \csc x | \bigg| \limits^{2 \pi}_{\pi} \Bigg)[/tex]Simplify:
[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \bigg( \ln | \csc x | - \ln | \sec x | + \frac{\sec^2 x - \csc^2 x}{2} \bigg) \bigg| \limits^{2 \pi}_{\pi}[/tex]Apply Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^{2 \pi}_{\pi} {\frac{\sin^6 x + \cos^6 x}{\sin^3 x \cos^3 x}} \, dx = \boxed{\text{un} \text{de} \text{f}}[/tex]

∴ we have found the value of the given integral.

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

Unit: Integration

Hello, can u help me out please i will give brainlist :)

Answers

Answer:

$5.75

Step-by-step explanation:

Find the difference between the weeks as this will give you the difference hence the amount deposited each week

0->1

is 12.00->17.75

diffrence is $5.75

1->2

is 17.75->23.50

diffrence is  $5.75

the amount saved each week is $5.75

What is x. Please show *all* the steps.

Answers

The equation 5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0 is a quadratic equation

The value of x is 8 or 1

How to determine the value of x?

The equation is given as:

5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Rewrite as:

-5/x - 2 + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Take the  LCM

[-5(x + 2) + (x -5)(x- 2)]\[x^2 - 4 + [3x + 8]/[x^2 - 4] = 0

Expand

[-5x - 10 + x^2 - 7x + 10]/[x^2 - 4] + [3x + 8]/[x^2 - 4] = 0

Evaluate the like terms

[x^2 - 12x]/[x^2 - 4] + [3x + 8]/[x^2 - 4 = 0

Multiply through by x^2 - 4

x^2 - 12x+ 3x + 8 = 0

Evaluate the like terms

x^2 -9x + 8 = 0

Expand

x^2 -x - 8x + 8 = 0

Factorize

x(x -1) - 8(x - 1) = 0

Factor out x - 1

(x -8)(x - 1) = 0

Solve for x

x = 8 or x = 1

Hence, the value of x is 8 or 1

Read more about equations at:

https://brainly.com/question/2972832

please someone do number 9

Answers

Answer:

2 4/5 kg.

Step-by-step explanation:

Multiply 7/2 (3 1/2) by 4/5

Bella is typing a 400 word essay. She types 50 words in 2 minutes.
Enter an equation in the form of y=mx+b, that represents the number of words (y) remaining to type after x minutes.

Answers

Answer:

[tex]y=25x[/tex]

Step-by-step explanation:

If Bella types 50 words in 2 minutes, we can divide 50 by 2 to see how many words she types in 1 minute.

[tex]50/2=25[/tex]

So, Bella types 25 words per minute. When she first began writing her essay, she did not have words to start with, so there is no y-intercept ([tex]b[/tex]).

Therefore, the equation that represents this relation is [tex]y=25x[/tex].

Solving the equation

We can divide 400 by 25 to determine how many minutes it would take her to finish her essay.

[tex]\frac{400}{25} =16[/tex]

Therefore, it will take her 16 minutes to complete her essay.

#LearnWithBrainly

1/15 ÷ 8 what is the answer to this question

Answers

Hey there!


1/15 ÷ 8

= 1/15 ÷ 8/1

= 1/15 × 1/8

= 1 × 1 / 15 × 8

= 1 / 120


Therefore, your answer is: 1/120


Good luck on your assignment & enjoy your day!


~Amphirite1040:)

someone please answer these 3 questions

Answers

Answer:

5 prism all have same volume

Step-by-step explanation:

Which answer choice correctly represents 0.4121212…?
A. 0.412
B.
C.
D.

Answers

Answer = the value in the second picture or middle

The one in the second picture
The bar on top of the numbers 12 means that the 12 keeps on re-occurring

the on with the line over the 2 only

Step-by-step explanation:

i do not know this answer

Answers

Answer:

9 millimeters is the answer.

6 a =24
Help me please

Answers

A would equals 4 because you would isolate the variable by dividing both sides by 6
I hope that I helped

kevin completed 10 problems of his homework during lunch this is 2/7 of the total assignment how many problem was Kevin assigned for homework

Answers

Answer:

35 problems

Step-by-step explanation:

We know that 10 problems are 2/7 of his total assignment, so we can set up the following equation,

[tex]\frac{10}{x\\} = \frac{2}{7}[/tex]

(side note: x is the denominator because we don't know how many problems it is out of yet)

We can cross multiply which gives us: (7x10) and 2(x)

70 = 2x (set them equal)

x = 35 (divide 70 by 2)

Hope this helps!

Solve for x. Leave your answer in simplest radical form.

SOMEONE PLEASE HELP ME ILL GIVE YOU BRAINLIST ANSWER

Answers

Answer:

x = 7

Step-by-step explanation:

I used the formula c^2 = a^2 + b^2

if you need further explanation,let me know

Scott has 6 T-shirts and 10 dress shirts in his wardrobe. What is the probability of selecting a dress shirt from his wardrobe?

a-3/8
b-3/5
c-5/8
d-8/5

Answers

Answer:

B

Step-by-step explanation:

6 divided by 10 = 0.6

0.6 = 6/10

6/10 = 3/5

The circumference of a circular ice
skating rink is 408.2 m.
What is the
approximate are of the skating rink? Use
3.14 for Pi. (10 points)

Answers

Answer:

13259.77 meters squared

Step-by-step explanation:

Generally, when finding the circumference of a circle, then formula is c=2*3.14*r (where r is the radius)

Substitute the given values

408.2= 2*3.14*r

R is 65

Now that we know that the radius is 65, the formula for the area of a circle is a=3.14*r^2

Substitute the given values

3.14*65^2

Therefore, the area of the circle is 13266.5 metres squared

ACTIVITY 2: Apply Pythagorean Theorem to complete the data below and write your answers on your answer sheet. Refer to the given examples. (2 points each number) A 1. 8 15 b 2. 18 24 3. 24 40 a C B 4. 16 20 5. 4 9 a² + b2 or Formulas: If c is unknown: c2 = If a is unknown: a2 = Ifb is unknown: b2 = c² - b2 or c= Va2 + b 2 a= vc2-b2 b = c2 - a2 = a² C2 - a2 or​

Answers

To completely complete the data use the formula a2 + b2 = c2 in order to find the value of the missing side.

What is the pytagorean theorem?

This is a general formula that can be used to find out the lenght of one of the sides in a right-angle triangle by knowing the other sides.

The general formula is:

a^2 + b^2 = c^2side a = base of the triangleside b = altitudeside c= hipotenuse (diagonal line)

How to use the theorem?

Replace the known sides and find the third side. For example:

a^2 + b^2 = c^25^2+6^2 = c^215 + 36 = c^251 = c^2c = √51c = 7.14

Learn more about Pythagorean Theorem in: https://brainly.com/question/14504604

Is 2x + 6x equivalent to 9x when x=0?

Answers

Answer:

Yes

Step-by-step explanation:

Anything muliplied by zero end up equalling zero. So with this in mind we can put 0 in for x and find out what each eqautian equals.

2(0) + 6(0) = 0; Once we multiply both sides we end up adding 0 and 0, so it equals 0

9(0) = 0

Since, 0 = 0 they are both equal.

Hope this helps!

What are the solutions to the quadratic equation (x + 4)(3x - 5) = 0?

Answers

Answer:

x = 4x = 5/3

Step-by-step explanation:

Given :

⇒ (x + 4)(3x - 5) = 0

The solutions will make the answer 0.

Therefore,

x + 4 = 0 3x - 5 = 0

⇒ x = 4

⇒ 3x = 5 ⇒ x = 5/3

a water tank holds 256 gallons but is leaking at 3 gallons per week. A second water tank holds 384 gallons but is leaking at a rate of 5 gallons per week . After how many weeks will the amount of water in the two tanks be the same

Answers

the two tanks wi equal 64

x=64

Answer:

w = 64 weeks

Step-by-step explanation:

From the information, we have two equations that we will need to set equal to each other to find the number of weeks, w,  it takes for both tanks to hold the same amount:  

[tex]-3w+256=-5w+384\\(-3w+256=-5w+384)-256\\(-3w=-5w+128)+5w\\(2w=128) /2\\w=64[/tex]

Determine if the two triangles are necessarily congruent. If so, fill in a flowchart proof to prove that they are. N A B M​

Answers

Answer:

Step-by-step explanation:

There are 25 white tiles in a box. what percent of tiles what percent of tiles will ally use to tile her laundry room floor (6ft by 6ft)

Answers

Answer: 100%

Step-by-step explanation:

Find the value of x. (25 points!)

Answers

Answer:

22.5

Step-by-step explanation:

By the angle bisectors theorem,

30/32 = x/24

x=24(30/32) =22.5

Which expressions are equivalent to In e?

Answers

Answer:

lne = log(e)e = 1

Step-by-step explanation:

I hope this helps!

A tank in the shape of a hemisphere has a radius of 3 feet. If the liquid that fills the tank has a density of 88.2 pounds per cubic foot, what is the total weight of the liquid in the tank, to the nearest full pound?

Answers

Answer:

4988 pounds

Step-by-step explanation:

Volume of Hemisphere

4/3 x π x r^3 = 4/3 x 3^3 x π = 4/3 x 27π = 36π = Volume of Sphere

To calculate volume of Hemisphere = 36π / 2 = 18π foot^3

Weight of the liquid

88.2 pounds / foot^3 means that every cubic foot weighs 88.2 pounds so

88.2 pounds x 18π = 4987.5925 which to the nearest pound = 4988 pounds.

What is the value of c in the equation below?
2^-4 x 2^2 = a^b = c

Answers

Answer:

[tex]\frac{1}{64}[/tex]

The third option.

Step-by-step explanation:

Use DeMoivre's Theorem to find the square roots of 4(cos90+ i sin 90). Show your work.

Answers

Answer:

Procedure 6.3. 1:Finding Roots of a Complex Number

Express both z and w in polar form z=reiθ,w=seiϕ. ...

Solve the following two equations: rn=s. ...

The solutions to rn=s are given by r=n√s.

The solutions to einθ=eiϕ are given by: nθ=ϕ+2πℓ,forℓ=0,1,2,⋯,n−1. ...

Using the solutions r,θ to the equations given in (6.3.

Step-by-step explanation:

Answer:-1.79229446 + 3.57598665 i

Step-by-step explanation:

I will honest this how much I can help you

The Ferris wheel shown makes 14 revolutions per ride. How far would someone travel during one ride? Round your final answer to the nearest whole number. Use 3.14 for π.


diameter 67 feet


Someone would travel about
feet during one ride on the Ferris wheel.

Answers

Answer:

2,945 feet

Step-by-step explanation:

Find the circumference.

C = dπ

C = 67(3.14)

C = 210.38

Since there are 14 revolutions, multiply the circumference by 14.

210.38 x 14 = 2945.32

Round

2,945

Find the solution set of the inequality
4.7 -1 <11.

Answers

Answer:

4x-1<11

4x<11+1

4x<12

x<12/4

x<3

I need help with Function Operations please

Answers

Answer:

  30

Step-by-step explanation:

The ring operator signifies a composition of functions. The composition is evaluated right-to-left. That means the composition ...

  [tex](f\circ g)(x)[/tex]

should be interpreted to mean ...

  [tex]f(g(x))[/tex]

___

To evaluate f(g(6)), we first evaluate g(6):

  g(6) = 6² -5 = 31

Then we evaluate the function f using that as its argument:

  f(31) = 31 -1 = 30

  f(g(x)) = 30

A radioactive compound with mass of 260 grams decays at a rate of 3.8% per hour. Which equation represents how many grams of the compound will remain after 8 hours?

A: C=260(1.038)^8C=260(1.038)^8

B: C=260(0.962)^8C=260(0.962)^8

C: C=260(0.038)^8C=260(0.038)^8

D: C=260(1-0.38)^8C=260(1−0.38)^8

Answers

The exponential equation that represents how many grams of the compound will remain after 8 hours is:

B: C=260(0.962)^8

What is an exponential function?

A decaying exponential function is modeled by:

[tex]A(t) = A(0)(1 - r)^t[/tex]

In which:

A(0) is the initial value.r is the decay rate, as a decimal.

A radioactive compound with mass of 260 grams decays at a rate of 3.8% per hour, hence the equation for the mass after t hours is given by:

[tex]A(t) = 260(1 - 0.038)^t[/tex]

[tex]A(t) = 260(0.962)^t[/tex]

After 8 hours, the amount is given by:

[tex]A(t) = 260(0.962)^8[/tex]

Hence option B is correct.

More can be learned about exponential equations at https://brainly.com/question/25537936

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