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Y_Kistochka [10]
3 years ago
6

7. Juan borrows $115 from his parents for a new cell phone. He agrees to pay his parents back the

Mathematics
1 answer:
Butoxors [25]3 years ago
5 0
I think the answer would be $119.60.

you start by doing 115*.02=2.3. this means that every year, juan will have to pay an extra $2.3. you then multiply 2.3 by 2, since it’s taking 2 years, getting 4.6. if you do 115+4.6, you get your total of $119.60. im pretty sure that should be correct. :-)
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Llana [10]
7, 13 , 19.

An arithmetic sequence is one that has a common difference.

13 - 7 = 6.
19 - 13 = 6.

So the next in the sequence is 19 + 6 = 25.

The next in the sequence is 25. 

It is not affected by what any one says. 
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Yu, Nailah, and Elena each bought between 7 and 9 yards of ribbon. Yu bought 3 pieces of ribbon. Nailah bought 5 pieces of ribbo
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6 0
3 years ago
An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
exis [7]
Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
0 = -w \sin(wt) + \sqrt{8} w cos(wt)
divide by w
0 =- \sin(wt) + \sqrt{8} cos(wt)
we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 

substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}
so the maximum value =\frac{1}{3}
minimum value =- \frac{1}{3}


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3 years ago
Brainliest
Oduvanchick [21]

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Step-by-step explanation:

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