A pile of sand has been stored in the shape of a cone. Mr. Terwilliker knows that the pile is 20 feet tall and 102 feet in circumference at the base. What is the area of the conical tarpaulin needed to cover the pile?

Answer:

8x

Step-by-step explanation:

=3x+5x

=8x

Answer:

c

Step-by-step explanation:

Answer:

side A should be about 44cm

Answer:

Step-by-step explanation:

Answer:

8-x=2

-x=2-8

-x=-6

x=6

if 8 is subtract from 6answer is 2

Answer:

Step-by-step explanation:

The series is missing from the question. I will answer this question with a general explanation by using the following similar series:

[tex]\sum\limits^6_{n=0} 7^n[/tex]

Required

The series

To do this, we simply replace n with the values

[tex]\sum\limits^6_{n=0}[/tex] means n starts from 0 and ends at 6

[tex]\sum[/tex] means the series is a summation series

So, we have:

[tex]\sum\limits^6_{n=0} 7^n = 7^0 + 7^1 + 7^2 + ...... + 7^6[/tex]

[tex]\sum\limits^6_{n=0} 7^n = 1 + 7 + 7^2 + ...... + 7^6[/tex]

Answer:

hey the answer is cylinder= 211.5л!

Answer:

Answer:

108 squared centimeters

Step-by-step explanation:

Let's exchange π for 3:

area = πr^2

        = 3r^2

Now, as you can see, the radius of this circle is 6. Let's plug in the value of r:

area = 3r^2

        = 3 · 6^2

Simplify 6^2:

area = 3 · 6^2

        = 3 · 36

Multiply 3 by 36:

area = 3 · 36

area = 108

108 squared centimeters

Answer:

The p-value of the test is 0.166 > 0.05, which means that there is not sufficient evidence at the 0.05 significance level to conclude that the bags contain more than 23.6 ounces.

Step-by-step explanation:

A company distributes candies in bags labeled 23.6 ounces. Test if the mean is more than this:

At the null hypothesis, we test if the mean is of 23.6, that is:

[tex]H_0: \mu = 23.6[/tex]

At the alternative hypothesis, we test if the mean is of more than 23.6, that is:

[tex]H_1: \mu > 23.6[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

23.6 is tested at the null hypothesis:

This means that [tex]\mu = 23.6[/tex]

The local bureau of weights and Measures randomly selects 60 bags of candies and obtain a sample mean of 24 ounces. Assuming that the standard deviation is 3.2.

This means that [tex]n = 60, X = 24, \sigma = 3.2[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{24 - 23.6}{\frac{3.2}{\sqrt{60}}}[/tex]

[tex]z = 0.97[/tex]

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 24, which is 1 subtracted by the p-value of z = 0.97.

Looking at the z-table, z = 0.97 has a p-value of 0.834.

1 - 0.834 = 0.166

The p-value of the test is 0.166 > 0.05, which means that there is not sufficient evidence at the 0.05 significance level to conclude that the bags contain more than 23.6 ounces.

please mark this answer as brainlist

Answer:

8.16 or 6.06

Step-by-step explanation:

final is 3.96

they reduced their time twice by 2.1min

3.96+2.1+2. 1=8.16

Answer:

Shannon

Step-by-step explanation:

Cuando se habla de lo que uno tiene, podemos usar números positivos.

Por ejemplo:

Pedro tiene 10 manzanas.

Para el caso de deudas, utilizamos números negativos, por ejemplo:

Pedro tiene -10 manzanas

Lo cual significa que Pedro debe 10 manzanas a alguien.

Entonces si le diéramos a Pedro 12 manzanas, el ahora tendría:

-10 + 12 = 2

Pedro tiene 2 manzanas, porque tuvo que entregar 10 de las 12 que le dimos para pagar su deuda.

Ahora vamos a resolver el problema:

La cuenta de Sally tiene un saldo de:

S = -$200.90

El signo negativo quiere decir que Sally tiene una deuda de $200.90

La cuenta de su amiga Shannon tiene un saldo de:

S' = -$240.55

De vuelta, el signo negativo quiere decir que Shannon tiene una deuda de $240.55

Con esto ya podemos concluir que la deuda de Shannon es mayor, por lo tanto Shannon es la que tiene más deuda.

Answer:

-10k×2+7

= -20k+7

Step-by-step explanation:

is the answer

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

Work Shown:

Plug in x = 0

[tex]f(x) = \frac{2x-1}{3x+5}\\\\f(0) = \frac{2*0-1}{3*0+5}\\\\f(0) = \frac{0-1}{0+5}\\\\f(0) = -\frac{1}{5}\\\\[/tex]

Repeat for x = 8

[tex]f(x) = \frac{2x-1}{3x+5}\\\\f(8) = \frac{2*8-1}{3*8+5}\\\\f(8) = \frac{16-1}{24+5}\\\\f(8) = \frac{15}{29}\\\\[/tex]

Now use the average rate of change formula

[tex]m = \frac{f(b)-f(a)}{b-a}\\\\m = \frac{f(8)-f(0)}{8-0}\\\\m = \frac{15/29 - (-1/5)}{8}\\\\m = \frac{15/29 + 1/5}{8}\\\\m = \frac{(15/29)*(5/5) + (1/5)*(29/29)}{8}\\\\m = \frac{75/145 + 29/145}{8}\\\\[/tex]

[tex]m=\frac{104/145}{8}\\\\m = \frac{104}{145} \div \frac{8}{1}\\\\m = \frac{104}{145} \times \frac{1}{8}\\\\m = \frac{104*1}{145*8}\\\\m = \frac{104}{1160}\\\\m = \frac{13}{145}\\\\[/tex]

Answer:

[tex]{ \tt{(2x + y)( {n}^{2} - 3xy + {y}^{2} )}} \\ = { \tt{(2x {n}^{2} - 3 {x}^{2}y + 2x {y}^{2} + {n}^{2}y - 3x {y}^{2} + {y}^{3} )}} \\ = { \tt{ {y}^{3} - xy(y + 3x) + {n}^{2} y }}[/tex]

Hello!

(2x+y) (n2-3xy+y2)

2x* n²= 2xn²

2x* -3xy = -6x²y

2x* y² = 2xy²

y*n² = yn²

y*-3xy = -3xy²

y* y² = y³

=>

2xn²- 6x²y + 2xy² +yn²- 3xy² +y³

Option C

Corresponding angles along parrellel lines are conguerent

Answered by Gauthmath pls mark brainliest and comment thanks and click thanks

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]

Answer:

c 6m hope to help

tjabssb

Answer:

3 m

Step-by-step explanation:

because all sides are equal

Answer:

yes the x increases by 6 and the y decreases by 3.

y = -1/2x - 7/2

Step-by-step explanation:

find the slope :

(1,-4), (7, -7)

y2- y1 / x2 - x1

substitute those numbers and you get -1/2.

point slope form :

y - y1 = m(x- x1)

y - (-4) = -1/2 ( x - (1))

y+4 = -1/2(x-1)

slope intercept form :

y = -1/2x - 7/2

does this help ?

Step-by-step explanation:

A ball is thrown straight up from a rooftop 320 feet high. The formula below describes the ball's height above the ground, h, in feet, t seconds after it was thrown. The ball misses the rooftop on its way down and eventually strikes the ground. How long will it take for the ball to hit the ground? Use this information to provide tick marks with appropriate numbers along the horizontal axis in the figure shown.

h=-16t^2+16t+320

Answer:

[tex]10^{-3}[/tex]

Step-by-step explanation:

Answer:

https://tex.z-dn.net/?f=10%5E%7B-3%7D

Step-by-step explanation:

Answer:

a.) 19.2cm

b.) 0.15375cm

Step-by-step explanation:

Cylinders are similar, so:

h1 / r1 = h2 / r2

8cm / 5cm = h2 / 12cm

h2 = (8cm × 12cm) / 5cm

h2 = 19.2cm

Same for b

32000cm2 / 246cm = 20cm2 / length

length = ((20 × 246) / 32000) cm

length = 0.15375cm

Answer:

The function that represents g(x) is the third choice: g(x) = (x − 4)^2 + 9

Step-by-step explanation:

The original function has been shifted 9 units up (a vertical transformation). To show a vertical transformation, all we have to do is either add or subtract at the end of the function.

To show a shift upwards, we add the value of change.

To show a shift downwards, we subtract the value of change.

In this case, the original function f(x) = [tex]x^{2}[/tex] was translated 9 units up. Since we shifted up, we simply add 9 to the end of the function: g(x) = [tex]x^{2}[/tex] + 9

The original function has also been shifted 4 units to the right. This is a horizontal transformation. To show a horizontal transformation, we need to either add or subtract within the function (within the parenthesis).

To show a shift to the left, we add the value of change.

To show a shift to the right, we subtract the value of change.

*Notice: Moving left does NOT mean to subtract while moving right does NOT mean to add. The rules above are counterintuitive so pay attention when doing horizontal transformations.

In this case, the original function f(x) = [tex]x^{2}[/tex] was translated 4 units to the right. Since we shifted right, we must subtract 4 units within the function/parenthesis: g(x) = [tex](x-4)^{2}[/tex]

When we combine both vertical and horizontal changes, the only equation that follows these rules is the third choice: g(x) = (x − 4)^2 + 9

Answer: C

Step-by-step explanation: