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vitfil [10]
4 years ago
12

Please help! .!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
Bess [88]4 years ago
8 0

Answer:

(3, -1) is the answer to this problem.

miv72 [106K]4 years ago
8 0

Answer:

3,-1

Step-by-step explanation:

You have (x,y) and you can put in 3 as x and -1 for y

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Can anyone explain step by step how to get this answer?<br><br><br> Will mark branliest!!
Y_Kistochka [10]

Answer:

Step-by-step explanation:

28°

4 0
3 years ago
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Please help! will give brainliest:)
Lorico [155]
Answer

1) x = 8

2) x = 1/3y + -5/3z

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Hope this helps:))
7 0
3 years ago
Let p(x) = x^2 + 8x + 12 and q(x) = x^3 + 5x^2 - 6x. Define r(x) = p(x) \cdot q(x).
Anna [14]

The roots of the entire <em>polynomic</em> expression, that is, the product of p(x) = x^2 + 8x + 12 and q(x) = x^3 + 5x^2 - 6x, are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6.

<h3>How to solve a product of two polynomials </h3>

A value of <em>x</em> is said to be a root of the polynomial if and only if <em>r(x) =</em> 0. Let be <em>r(x) = p(x) · q(x)</em>, then we need to find the roots both for <em>p(x)</em> and <em>q(x)</em> by factoring each polynomial, the factoring is based on algebraic properties:

<em>r(x) =</em> (x + 6) · (x + 2) · x · (x² + 5 · x - 6)

<em>r(x) =</em> (x + 6) · (x + 2) · x · (x + 3) · (x + 2)

r(x) = x · (x + 2)² · (x + 3) · (x + 6)

By direct inspection, we conclude that the roots of the entire <em>polynomic</em> expression are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6. \blacksquare

To learn more on polynomials, we kindly invite to check this verified question: brainly.com/question/11536910

5 0
2 years ago
Complete the coordinate proof for the quadrilateral determined by the points A(7, 9), B(9, 4) , C(4, 2), D(2, 7) . Prove that AB
seraphim [82]

Answer:

its B

Step-by-step explanation:

8 0
3 years ago
Answer the question for 10 points
Fynjy0 [20]

Answer:

The answer to the mean is 66.4

Step-by-step explanation:

58+63+68+72+71

8 0
3 years ago
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