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1)  If \(11^{5y}=\left(\frac{1}{121}\right)^{2y+3}\), find \(y\). Express your answer in simplest fractional form.

 

2) It's January 1st, and Jillian is starting a new savings plan. Her savings account charges a \( \$10\) monthly fee for any month where she does not have an ending balance of\( \$1,\!000\) or higher. At the beginning of each month, she deposits \(\$100\) into her account. At the end of the year, her bank awards her an interest at the rate of \(1\%.\) rounded to the nearest dollar, calculated on the ending balance.

After three years, what is her ending balance after interest?

 

Thx in advance

-wolf

 Aug 29, 2019
 #1
avatar+104683 
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1)   11^(5y)   =  (1 /121)^(2y+3)          {Note : 1/121   =  1/11^2  = 11^(-2)   }

 

So we have

 

11^(5y)   =  (11^(-2) )^(2y + 3)

 

11^(5y)  =  (11)^(-4y - 6)        we have the bases the same......solve for the exponents

 

5y = -4y - 6      add 4y to both sides

 

9y  = -6              divide both sides by 9

 

y =  -6/9   =  -2/3

 

 

cool cool cool

 Aug 29, 2019
 #2
avatar+104683 
+1

At the end of October of the first year  she will have   (100 - 10)*10  = 90 * 10 =   $900 in her account

In November....she  contributes $100 and she will have  $1000 in her acct at the end of November

In December, she contributes $100  and she will have $1000 + $100  = $1100 in her acct at the end of the year

She  will then make 1% interest on this  so she will have an ending balance of 1100 (1.01)  = $1111   in her account

 

The second year  she adds  $1200  to this  = $2311  and she makes 1% interest on this  = 2311 (1.01)  =

$2334 at the end of  the second year

 

And in the third year...she adds $1200 more to this  = 3534 and she makes 1% interest on this and ends up with

3534(1.01) = $3569

 

CORRECTED ANSWER.......!!!!

 

cool cool cool

 Aug 29, 2019
edited by CPhill  Aug 29, 2019
 #3
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Thx for the help but question 2 is incorrect, Thx for quesiton 1 tho laugh

 

-wolf

 Aug 29, 2019
 #4
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Nvm I fot the correct answer now

Guest Aug 29, 2019

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