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Ira Lisetskai [31]
3 years ago
6

Michael had $550 in his savings account. He took out $30.75 every month for one year. What is the net change in Michael’s accoun

t balance following these withdrawals?
A.
-$519.25
B.
-$369
C.
$369
D.
$519.25
Mathematics
2 answers:
gulaghasi [49]3 years ago
5 0

Answer:

B

Step-by-step explanation:

Aleks04 [339]3 years ago
4 0

I believe it to be $369

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Find the fifth term in the sequence<br>that is defined as follows:<br>an = 2n​
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Answer:

a_{5}=10

Step-by-step explanation:

If you want to find the 5th term of a sequence or any term you have to substitute n for the desired term in your case the result is the following

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3 years ago
The School has budgeted $3,500 for an end- of -year party in the local park. The cost to rent the park shelter is $250. How much
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Answer:

$5.3

Step-by-step explanation:

<u>1) Ask yourself: How much money is left to spend? </u>

<em>- Subtract the cost to rent the park shelter: </em>

3,500 - 250 = 3,250

<em>- Find the total amount of money spent on gifts for students:   </em>

300 * 5.50 = 1,650  

<em>- Subtract the total money spent on gifts for students: </em>

3,250 - 1,650 = 1,600

<u>2) Ask yourself: How much can be spent </u><u>per student</u><u>? </u>

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1,600 / 300 = 5.3

7 0
2 years ago
Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
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Answer:

\displaystyle  -\frac{1}{2} \leq x < 1

Step-by-step explanation:

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We must find the set of values for x that make the expression stand

\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

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Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

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(6x^3+14x^2+10x+12)

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