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2 years ago

5. I have 25 coins consisting of 5-cent and 50-cent pieces. If the total value is $7.10, how many 5-cent coins do I

Mathematics

1 answer:

VARVARA [1.3K]2 years ago

6 0

Answer:

x = # of 5 cent coins

y = # of 50 cent coins

x+y=25

.05x+.5y=7.1

y=25-x

(.05x+.5(25-x)=7.1)100

5x+50(25-x)=710

5x+1250-50x=710

-45x=-540

x=12

You have 12, 5 cent coins and 13, 50 cent coins.

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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!

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3 years ago

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IceJOKER [234]

Answer:

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Step-by-step explanation:

The smallest the third side can be is greater than the difference of the other two sides

x > 5.6 -4

x > 1.6

The largest the third side can be is less than the sum of the other two sides

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