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olga_2 [115]
3 years ago
6

Find surface area of the prism

Mathematics
1 answer:
Vika [28.1K]3 years ago
6 0
SA=(a+b+c) h+bh
=(5+12+13)*4+5*12
=20*4+60
=80+60
=140 cm squared
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PLS HELP I NeED to finish this
mote1985 [20]

Answer:

D. 15

Step-by-step explanation:

<em>Use proportions.</em>

<em />\frac{35}{25} = \frac{21}{x}  \\\\35x = 525\\x = 15<em />

3 0
3 years ago
Find the length of the unknown side round your answer to the nearest tenth
iragen [17]

Answer:

C

Step-by-step explanation:

4 0
3 years ago
Sqaure root of 5 (u-1)(u^5+u^4+u^3+u^2+u+1)
Anna71 [15]
GIVEN:

5(u - 1)( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1)

remember:

\sqrt{u} = {u}^{ \frac{1}{2} }

And

{u}^{n} \times {u}^{m} = {u}^{n + m}

SOLVE:

start by multiplying the factors:

5( ({u}^{6} + {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u ) - ( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1))

simplify by combing like terms. Most terms subtract off, leaving:

5( {u}^{6} - 1)

This can be factored, but it is not a perfect square, which is really what we need to take the square root.

5( {u}^{3} - 1)( {u}^{3} + 1)

I'm not exactly sure what form they want the answer in...

so taking the square root:

\sqrt{5( {u}^{6} - 1) } = {(5( {u}^{6} - 1))}^{ \frac{1}{2} }

so my best answer is:

{5}^{ \frac{1}{2} } \times {( {u}^{6} - 1)}^{ \frac{1}{2} }
or the more factored form:

{5}^{ \frac{1}{2} } { ({u}^{3} - 1)}^{ \frac{1}{2} } { ({u}^{3} + 1 )}^{ \frac{1}{2} }

I'm not sure how else to solve it. Taking the square root doesn't work out super well, so I left it in the most simple form I could.

sorry for not coming to a definitive answer!
7 0
3 years ago
Joshua's Cousin Pledged 12$ If Joshua's walked 3 miles how much did his cousin Pay per mile
Alexus [3.1K]
4$
each mile us 4$
12÷3=4$
7 0
4 years ago
Read 2 more answers
The graph shows calories compared to grams of protein.
andrezito [222]

Hello :3

Your answer would be C. (or "There are 12 calories in 3 grams of protein).

Hope This Helps (Hope i am not to late sorry if i am)

Cupkake~

3 0
3 years ago
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