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snow_tiger [21]
3 years ago
9

Solve the system by submission -5y-10=x x+9y=-2

Mathematics
1 answer:
Sonbull [250]3 years ago
8 0

Answer:  =−20

y=2

​

Step-by-step explanation:

Solve for xx in -5y-10=x−5y−10=x.

x=-5y-10

x=−5y−10

2 Substitute x=-5y-10x=−5y−10 into x+9y=-2x+9y=−2.

4y-10=-2

4y−10=−2

3 Solve for yy in 4y-10=-24y−10=−2.

y=2

y=2

4 Substitute y=2y=2 into x=-5y-10x=−5y−10.

x=-20

x=−20

5 Therefore,

\begin{aligned}&x=-20\\&y=2\end{aligned}

​

 

x=−20

y=2

​

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What is the area of the front face?
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the area of the front face would be 40 square inches.

Step-by-step explanation:

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7 0
3 years ago
What are the zeros of the quadratic function f(x) = 6x2 + 12x - 7?<br>​
mina [271]

Answer:

\frac{-6+\sqrt{78} }{6} and \frac{-6-\sqrt{78} }{6}

Step-by-step explanation:

We are asked that what are the zeros of the quadratic function f(x) = 6x² +12x -7.

So, we have to find the roots of the equation, f(x) = 6x² +12x -7 =0 ...... (1)

Since the quadratic function can not be factorized, so we have to apply Sridhar Acharya's formula.

This formula gives if, ax² +bx +c =0, the the two roots of the equation are

\frac{-b+\sqrt{b^{2}-4ac } }{2a} , \frac{-b-\sqrt{b^{2}-4ac } }{2a}

Therefore, in our case 'a' being 6, 'b' being 12 and 'c' being -7, the two roots of the equation (1) will be

\frac{-12+\sqrt{12^{2}-4*6*(-7) } }{2*6},  \frac{-12-\sqrt{12^{2}-4*6*(-7) } }{2*6}

= \frac{-12+\sqrt{312} }{12},\frac{-12-\sqrt{312} }{12}

=\frac{-6+\sqrt{78} }{6} and \frac{-6-\sqrt{78} }{6}

Hence, x= \frac{-6+\sqrt{78} }{6}

and x= \frac{-6-\sqrt{78} }{6}

(Answer)

9 0
4 years ago
Plssssssssssssssssssssss help
bearhunter [10]

Answer:

10-8

9-9

11-7

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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