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blondinia [14]
3 years ago
7

In your notebook, set up the following addition using a vertical format and find the sum of the given polynomials. 3x^2 + 2x - 5

and -4 + 7x^2
-x 2 + 9x - 9
21x 2 + 2x + 20
10x 2 - 5x - 1
10x 2 + 2x - 9
ANSWER ASAP please
Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
5 0
The sum to your question is 14
e-lub [12.9K]3 years ago
4 0

Answer #1- Add the expressions.  3x^2+2x−5+(−4+7x^2), Remove unnecessary parentheses.  3x^2+2x−5−4+7x^2, Add  3x^2 and 7x^2. 10x^2+2x−5−4, Subtract  4  from  −5. 10x^2+2x−9.

Final Answer: 10x^2  +2x−9 hopefully you now get an understanding of how to do it.

Answer #2: 7x−9

Answer #3: 4(11x+5)

Answer #4: 15x-1

Answer #5: 22x-9

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Answer:

189 chips

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8 0
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mash [69]

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123

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5 0
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Softa [21]
180^o-112^o=68^o

\left\{\begin{array}{ccc}2x+3y=68&|\cdot(-15)\\30x+11y=112\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-30x-45y=-1020\\30x+11y=112\end{array}\right}\\.\ \ \ \ \ \ \ -34y=-908\ \ \ |:(-34)\\\\.\ \ \ \ \ \ \ \ \ \ \  y=\dfrac{908}{34}\to y=\dfrac{454}{17}\\\\2x+3\cdot\dfrac{454}{17}=68\ \ \ \ |\cdot17\\\\34x+1362=1156\ \ \ \ |-1362\\34x=-206\ \ \ \ |:34\\\\x=-\dfrac{206}{34}\to x=-\dfrac{103}{17}\\\\  \left\{\begin{array}{ccc}x=-\dfrac{103}{17}\\\\y=\dfrac{454}{17}\end{array}\right
8 0
3 years ago
Deangelo has $7 worth of dimes and quarters in a jar. He has 7 more quarters than dimes.
svlad2 [7]
D=number of dimes
q=number of quarters

let's count everything in cents
dimes are worth 10 cents
quarters are worth 25 cents

10d+25q=700
divide both sides by 5
2d+5q=140


he has 7 more quarters than dimes
q=7+d
subsitute 7+d for q in other equation

2d+5q=140
2d+5(7+d)=140
2d+35+5d=140
7d+35=140
minus 35 both sides
7d=105
divide both sides by 7
d=15

sub back

q=7+d
q=7+15
q=22


22 quarters and 15 dimes
3 0
3 years ago
Read 2 more answers
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