Ask question

Ask question

All categories

2 years ago

9

A cylinder, cone, and sphere are shown below. The three figures have the same radius. The cylinder and cone have the same height

with h = r. Use this information to answer question

1 answer:

Mice21 [21]2 years ago

8 0

I’m just trying so much better to talk now but it’s just a good 256

You might be interested in

A set of equations is given below:

Hunter-Best [27]

If you would like to find a step that can be used to find the solution to the set of equations, you can do this using the following steps:

c = 2d + 1
c = 3d + 5
__________
c = c
2d + 1 = 3d + 5
2d - 3d = 5 - 1
- d = 4
d = -4

The correct result would be <span>2d + 1 = 3d + 5.</span>

8 0

3 years ago

What is the means-to-MAD ratio of the two data sets, expressed as a decimal to the nearest tenth?

FinnZ [79.3K]

Frist we need to find the mean of both so

1.55

2.67

then we make a number graph and see how mant place does it take to get to them from 0 67and55/by them selfs = 1.2(rounded)

1.2 is you answer

4 0

3 years ago

Use the figure to the find the measures of the numbered angles

mina [271]

Answer:

sdcce

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

YO!!!!WILL GIVE THE BRAIN!!!!!!!!

Tatiana [17]

The answer is b, since that's the only other function there that involves using absolute value

3 0

3 years ago

Read 2 more answers

Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.

Leya [2.2K]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 2}\implies A=6500(1.12)^2\implies A=8153.6$

6 0

3 years ago