The circumference of a circle is 6pi inches. What is the area of the circle

Answer:

7t + 6 + 9v

Step-by-step explanation:

7t + 6 + 3v + 6v (since 3v and 6v are like terms you will add them both.)

7t + 6 + 9v

Hope this helps, thank you :) !!

Answer:

7t+6+9v

Step-by-step explanation:

7t+6+3v+6v

7t has no opponent it is =7t

6 is on it own =6

3v+6v=9v,reason is 3v has an opponent which is 6v so addition of 3v and 6v is =9v

so ur ans. is =7t+6+9v

Answer:

6

BRAINLIEST, PLEASE!

Step-by-step explanation:

-19p - 2p + 16p + 12 = -18

-5p + 12 = -18

-5p = -30

p = 6

Answer:

p = 6

Step-by-step explanation:

Given

- 19p - 2p + 16p + 12 = - 18 ( simplify left side )

- 5p + 12 = - 18 ( subtract 12 from both sides )

- 5p = - 30 ( divide both sides by - 5 )

p = 6

Answer:

Slope is (1/4)

Step-by-step explanation:

The slope is calculated by (6-5)/(5-1)=1/4

The type of budget that would likely work best for Brody is biweeky budget.

Budget is an economic term that refers to the planning and advance formulation of expenses and income. The budget is a tool to organize expenses depending on the amount of money available.

The type of budget that would be best for Brody is a biweekly budget because he receives his payment every fifteen days (twice a month). So, he  can schedule his expenses each time he receives his payment, in this way he  does not spend all his money before he receives the next payment.

Additionally, weekly, monthly, and dairy are not correct options because they do not fit the time periods in which Brody receives payment for his services.

Learn more in: https://brainly.com/question/141889

Note:

This question is incomplete because options are missing, here are the options.

Daily budget

Biweekly budget

Monthly budget

Weekly budget

Answer:

y = 10

Step-by-step explanation:

To find the y coordinate of the midpoint, take the y coordinates of the endpoints and average

(0+y)/2 = 5

Multiply each  die by 2

0+y = 10

y = 10

Answer: x= 11√3= 19.0525 = 19.1

Step-by-step explanation:

Let the reference angle be 30

so

cos 30 = b/h

√3/2 = x/22

or, 22√3 = 2x

or. x = (22√3)/2

so, x = 11√3

Answer:

x = 19.1 cm

Step-by-step explanation:

→ Find the name of the side you are not given

Opposite

→ Find a formula without opposite in it

Cos = Adjacent ÷ Hypotenuse

→ Rearrange to make adjacent the subject

Adjacent = Cos × Hypotenuse

→ Substitute in the values

Adjacent = Cos ( 30 ) × 22

→ Simplify

Adjacent = 19.1

Answer:

x(ax-1)-y(y+1)

Step-by-step explanation:

you have to group the like terms

ax^2-x-y^2-y

x(ax-1)-y(y+1)

I hope this helps

Basically count/add up the total amount of degrees that are include in the angle <FHD.

-- (central angles)

So, 35 + 65 = 100 degrees

Answer: g(x) = (1/2)3^-x  reflection over y axis yields (-x,y)

Answer:

216

Step-by-step explanation:

Volume of a rectangular prism = area of base * depth

Area of base: 72

Depth: 3

Volume = 72 * 3 = 216

Answer:

hope it help you

Step-by-step explanation:

mark me brailiest answer

Answer:

? ≈ 37°

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos? = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{4}{5}[/tex] , then

? = [tex]cos^{-1}[/tex] ([tex]\frac{4}{5}[/tex] ) ≈ 37° ( to the nearest degree )

Answer:

8n-2 = 6n+9

2n-2 = 9

2n = 11

n = 5.5

So C is correct

Let me know if this helps!

Answer:

divide, 2x+9

Step-by-step explanation:

got it right

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]\frac{1}{6}[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{Box Probability}}\\\\\rightarrow \frac{\text{# of boxed flawed}}{\text{# of boxes checked}} \\\\\rightarrow \frac{3}{18}\\\\\rightarrow \frac{3/3}{18/3}\\\\\rightarrow\boxed{\frac{1}{6}}[/tex]

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

149 inches squared

Step-by-step explanation:

top rectangle: 25 * 7 = 175

second rectangle: 8 * (25 - 17) = 8^2 = 64

triangle in bottom right: 1/2 * (13 - 8) * (15 - 11) = 10

175 + 64 + 10 = 149 sq in

hopefully got this right!

Answer:

9 years older

Step-by-step explanation:

The ratio of their ages is 8 : 5 = 8x : 5x ( x is a multiplier )

In 30 years their ages will be 8x + 30 and 5x + 30 and the ratio 6 : 5 , so

[tex]\frac{8x+30}{5x+30}[/tex] = [tex]\frac{6}{5}[/tex] ( cross- multiply )

5(8x + 30) = 6(5x + 30) ← distribute parenthesis on both sides

40x + 150 = 30x + 180 ( subtract 30x from both sides )

10x + 150 = 180 ( subtract 150 from both sides )

10x = 30 ( divide both sides by 10 )

x = 3

Then

Michell is 8x = 8 × 3 = 24 years old

Connor is 5x = 5 × 3 = 15 years old

Mitchell is 24 - 15 = 9 years older than Connor

9514 1404 393

Answer:

  equals can be substituted anytime anywhere

Step-by-step explanation:

cos(60°) = 1/2, so wherever one appears, the other can be substituted. This is allowed by the substitution property of equality.

__

If you don't substitute at some point, you find the answer to be ...

  x = 10/cos(60°)

Most of us are interested in a numerical value for x, so we prefer that cos(60°) be replaced by a numerical value.

Answer:

(-3)

Step-by-step explanation:

follow me if you want

a) 26 x (-48) + (-48) x (-36) = ( –1248) + ( + 1728) = – 1248+ 1728 = 480

b) 625 x (-35) + (-625) x 65 = ( –21875) + ( –40625) = – 21875 –40625 = –62500

I hope I helped you^_^

Answer:

See Below (Boxed Solutions).

Step-by-step explanation:

We are given the two complex numbers:

[tex]\displaystyle z = \sqrt{3} - i\text{ and } w = 6\left(\cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12}\right)[/tex]

First, convert z to polar form. Recall that polar form of a complex number is:

[tex]z=r\left(\cos \theta + i\sin\theta\right)[/tex]

We will first find its modulus r, which is given by:

[tex]\displaystyle r = |z| = \sqrt{a^2+b^2}[/tex]

In this case, a = √3 and b = -1. Thus, the modulus is:

[tex]r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2[/tex]

Next, find the argument θ in [0, 2π). Recall that:

[tex]\displaystyle \tan \theta = \frac{b}{a}[/tex]

Therefore:

[tex]\displaystyle \theta = \arctan\frac{(-1)}{\sqrt{3}}[/tex]

Evaluate:

[tex]\displaystyle \theta = -\frac{\pi}{6}[/tex]

Since z must be in QIV, using reference angles, the argument will be:

[tex]\displaystyle \theta = \frac{11\pi}{6}[/tex]

Therefore, z in polar form is:

[tex]\displaystyle z=2\left(\cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6}\right)[/tex]

Part A)

Recall that when multiplying two complex numbers z and w:

[tex]zw=r_1\cdot r_2 \left(\cos (\theta _1 + \theta _2) + i\sin(\theta_1 + \theta_2)\right)[/tex]

Therefore:

[tex]\displaystyle zw = (2)(6)\left(\cos\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right) + i\sin\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right)\right)[/tex]

Simplify. Hence, our polar form is:

[tex]\displaystyle\boxed{zw = 12\left(\cos\frac{9\pi}{4} + i\sin \frac{9\pi}{4}\right)}[/tex]

To find the complex form, evaluate:

[tex]\displaystyle zw = 12\cos \frac{9\pi}{4} + i\left(12\sin \frac{9\pi}{4}\right) =\boxed{ 6\sqrt{2} + 6i\sqrt{2}}[/tex]

Part B)

Recall that when raising a complex number to an exponent n:

[tex]\displaystyle z^n = r^n\left(\cos (n\cdot \theta) + i\sin (n\cdot \theta)\right)[/tex]

Therefore:

[tex]\displaystyle z^{10} = r^{10} \left(\cos (10\theta) + i\sin (10\theta)\right)[/tex]

Substitute:

[tex]\displaystyle z^{10} = (2)^{10} \left(\cos \left(10\left(\frac{11\pi}{6}\right)\right) + i\sin \left(10\left(\frac{11\pi}{6}\right)\right)\right)[/tex]

Simplify:

Simplify using coterminal angles. Thus, the polar form is:

[tex]\displaystyle \boxed{z^{10} = 1024\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)}[/tex]

And the complex form is:

[tex]\displaystyle z^{10} = 1024\cos \frac{\pi}{3} + i\left(1024\sin \frac{\pi}{3}\right) = \boxed{512+512i\sqrt{3}}[/tex]

Part C)

Recall that:

[tex]\displaystyle \frac{z}{w} = \frac{r_1}{r_2} \left(\cos (\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right)[/tex]

Therefore:

[tex]\displaystyle \frac{z}{w} = \frac{(2)}{(6)}\left(\cos \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right) + i \sin \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right)\right)[/tex]

Simplify. Hence, our polar form is:

[tex]\displaystyle\boxed{ \frac{z}{w} = \frac{1}{3} \left(\cos \frac{17\pi}{12} + i \sin \frac{17\pi}{12}\right)}[/tex]

And the complex form is:

[tex]\displaystyle \begin{aligned} \frac{z}{w} &= \frac{1}{3} \cos\frac{5\pi}{12} + i \left(\frac{1}{3} \sin \frac{5\pi}{12}\right)\right)\\ \\ &=\frac{1}{3}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right) + i\left(\frac{1}{3}\left(- \frac{\sqrt{6} + \sqrt{2}}{4}\right)\right) \\ \\ &= \boxed{\frac{\sqrt{2} - \sqrt{6}}{12} -\frac{\sqrt{6}+\sqrt{2}}{12}i}\end{aligned}[/tex]

Part D)

Let a be a cube root of z. Then by definition:

[tex]\displaystyle a^3 = z = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]

From the property in Part B, we know that:

[tex]\displaystyle a^3 = r^3\left(\cos (3\theta) + i\sin(3\theta)\right)[/tex]

Therefore:

[tex]\displaystyle r^3\left(\cos (3\theta) + i\sin (3\theta)\right) = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]

If two complex numbers are equal, their modulus and arguments must be equivalent. Thus:

[tex]\displaystyle r^3 = 2\text{ and } 3\theta = \frac{11\pi}{6}[/tex]

The first equation can be easily solved:

[tex]r=\sqrt[3]{2}[/tex]

For the second equation, 3θ must equal 11π/6 and any other rotation. In other words:

[tex]\displaystyle 3\theta = \frac{11\pi}{6} + 2\pi n\text{ where } n\in \mathbb{Z}[/tex]

Solve for the argument:

[tex]\displaystyle \theta = \frac{11\pi}{18} + \frac{2n\pi}{3} \text{ where } n \in \mathbb{Z}[/tex]

There are three distinct solutions within [0, 2π):

[tex]\displaystyle \theta = \frac{11\pi}{18} , \frac{23\pi}{18}\text{ and } \frac{35\pi}{18}[/tex]

Hence, the three roots are:

[tex]\displaystyle a_1 = \sqrt[3]{2} \left(\cos\frac{11\pi}{18}+ \sin \frac{11\pi}{18}\right) \\ \\ \\ a_2 = \sqrt[3]{2} \left(\cos \frac{23\pi}{18} + i\sin\frac{23\pi}{18}\right) \\ \\ \\ a_3 = \sqrt[3]{2} \left(\cos \frac{35\pi}{18} + i\sin \frac{35\pi}{18}\right)[/tex]

Or, approximately:

[tex]\displaystyle\boxed{ a _ 1\approx -0.4309 + 1.1839i,} \\ \\ \boxed{a_2 \approx -0.8099-0.9652i,} \\ \\ \boxed{a_3\approx 1.2408-0.2188i}[/tex]

LOL

Answer:

D. The over 30s have a larger range and interquartile range than the under 30s

Step-by-step explanation:

In a data set, the range is the difference between the maximum and minimum. So, the range for under 30s is 20, while the range for over 30s is 24. Additionally, the interquartile range is the difference between Q3 and Q1. For a boxplot, Q3 is the line where the box ends and Q1 is the line where the box begins. Therefore, the IQR for the under 30s is 8, and the IQR for over 30s is 11. So, D must be correct.

Answer:

Step-by-step explanation:

[tex]\frac{sinxcos^3x-cos xsin^3x}{cos^42x-sin^42x} \\=\frac{sin x cos x(cos^2x-sin ^2 x)}{(cis^2 2x+sin^2 2x)(cos^2 2x-sin ^22x)} \\=\frac{2sin x cos x cos 2x}{2(1)(cos 4x)} \\=\frac{sin 2x cos 2x}{2 cos 4x} \\=\frac{2 sin 2x cos 2x}{4 cos 4x} \\=\frac{sin 4x}{4 cos 4x} \\=\frac{1}{4} tan 4x[/tex]

Answer:

252

Step-by-step explanation:

To be divisible by 3, it's digits have to add to a number that is a multiple of 3.

To be divisible by 4 its last 2 digits have to be divisible by 3.

So let's start with 1x1 which won't work because 1x1 is odd. so let's go to 2x2 and see what happens.

212 that's divisible by 4 but not 3

222 divisible by 3 but not 4

232 divisible by 4 but not 3

242 not divisible by either one.

252 I think this might be your answer

The digits add up to 9 which is a multiple of 3 and the last 2 digits are divisible by 4

Answer:

1320

Step-by-step explanation:

Use the formula for sum of series, s(a) = n/2(2a + (n-1)d)

The terms increase by 6, so d is 6

a is the first term, -56

n is the terms you want to find, 33

Plug in the numbers, 33/2 (2(-56)+(32)6)

Simplify into 33(80)/2 and you get 1320

Answer:

47.746, answer choice C

Step-by-step explanation:

47.746

Answer:

range

Step-by-step explanation:

Answer:

B. range.

Step-by-step explanation:

others are:

» Standard variation.

» Interquatile range.

» Quatiles, deciles and percentiles.

» variance.

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