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dem82 [27]
4 years ago
9

what kinds of face shapes are the results from cuts made from parallel and perpendicular to the bases of right rectangular prism

s and pyramids
Mathematics
1 answer:
zubka84 [21]4 years ago
6 0
A cross section is the 2D shape that fallouts from cutting a 3D with a plane.To classify the face shape from cuts made parallel and perpendicular to the bases of right dimensional figures, we need to know first what parallel cuts are. These cuts will form the shape of the base while perpendicular cuts will take the form of the lateral face. Cuts at an angle through the right rectangular prism or pyramid will make a parallelogram.
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What is the sum of the geometric series in which A1 = 4, r = 3, and An = 324?
koban [17]
The answer would be 328
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3 years ago
<img src="https://tex.z-dn.net/?f=2x%2B7-1%3D2x%2B7" id="TexFormula1" title="2x+7-1=2x+7" alt="2x+7-1=2x+7" align="absmiddle" cl
Goshia [24]

Answer:

<h2>-1</h2>

Step-by-step explanation:

2x + 7  - 1 = 2x + 7

=  > 2x - 1 = 2x  + 7 - 7

=  > 2x - 1 = 2x

=  >  - 1 = 2x - 2x

=  > x =  - 1

7 0
3 years ago
You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp
Katena32 [7]
Let's move like a crab, backwards some.

after 2 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\to &\$1000\\&#10;r=rate\to 3\%\to \frac{3}{100}\to &0.03\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annually, thus once}&#10;\end{array}\to &1\\&#10;t=years\to &2&#10;\end{cases}&#10;\\\\\\&#10;A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2

after 3 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\to &\$1000\\&#10;r=rate\to 3\%\to \frac{3}{100}\to &0.03\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annually, thus once}&#10;\end{array}\to &1\\&#10;t=years\to &3&#10;\end{cases}&#10;\\\\\\&#10;A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3

is that enough to pay the $1100?


now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000   <---- standard form.
8 0
3 years ago
The number of cars (c) in a parking lot increases when the parking fee (f) decreases.
Vitek1552 [10]

The answer is c maybe?
5 0
4 years ago
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bixtya [17]

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Hope this helps! :)

3 0
2 years ago
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