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Taya2010 [7]
2 years ago
14

Carlos Martin received a statement from his bank showing a balance of $56.75 as of March 15. His checkbook shows a balance of $8

7.37 as of March 20. The bank returned all the cancelled checks but two. One check was for $5.00 and the other was for $13.25. How much did Carlos deposit in his account between March 15 and March 20?
Mathematics
2 answers:
Alenkinab [10]2 years ago
8 0
Answer is $22.37:
You subtract $56.75 from $87.37,which equals $30.62, then you add the 2 checks that the bank didn't return and subtract that amount from the $30.62.

Elina [12.6K]2 years ago
3 0
Answer is 48.87

subtract the 87 and 56, then add the 13.25 and 5 to the 30.62. This is what I did. havent finished the test yet but I fairly sure its right. As the answer's I've found so far haven't been on the list :)
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4 0
3 years ago
What's 1.25 times 12
Juli2301 [7.4K]
The awnser is 15 
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7 0
3 years ago
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The ordered pairs (12,32), (x,40), (18,48), (21,y) represent a proportional relationship. Find the values of x and y.
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write a eassy about importance of English language in student life

8 0
3 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
7 0
3 years ago
Area of square rug? It’s inside is (5x+1)
kherson [118]
(5x+1)(5x+1)=25x+5x+5x+1=35x+1

The area of the square rug is (35x+1).

I may be wrong, I never was good at these types of problems.
5 0
3 years ago
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