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# Here are two math problems :)

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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?

-wolf

Aug 29, 2019

#1
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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

Aug 29, 2019
#2
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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

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

-wolf

Aug 29, 2019
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Nvm I fot the correct answer now

Guest Aug 29, 2019