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3 years ago

9

(w+6)(w+7) how to solve

2 answers:

Makovka662 [10]3 years ago

8 0

(w+6)(w+7) = 0

w(w) + w(7) + 6(w) + 6(7) = 0
w² + 7w + 6w + 42 = 0
w² + 13w + 42 = 0

w + 6 = 0
w = -6
<span>w² + 13w + 42 = 0
</span>-6² + 13(-6) + 42 = 0
36 - 78 + 42 =
78 - 78 = 0
0 = 0

w + 7 = 0
w = -7

w² + 13w + 42 = 0
-7² + 13(-7) + 42 = 0
49 - 91 + 42 = 0
91-91 = 0
0=0

Wewaii [24]3 years ago

6 0

(w+6)(w+7)
= w^2+ 7w+ 6w+ 6*7
= w^2+ (7w+ 6w)+ 42 (combine like terms)
= w^2+ 13w+ 42

The final answer is w^2+ 13w+ 42~

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Step-by-step explanation:

Given

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Required

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To calculate the perimeter of A B C, we sum up the sides of the triangle

i.e.

$Perimeter = A B + B\ C + A\ C$

AB is given in the question already.

So, we need to calculate B F and F C

To solve for BF, we consider triangle B F C

Using Pythagoras Theorem, we have:

$B\ C^2 = B\ F^2 + F\ C^2$

Substitute values for BF and FC;

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$B\ C^2 = 81 + 31.36$

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$B\ C^2= \sqrt{112.36$

$B\ C = 10.6$

Next, we calculate A C.

To calculate A C, we need to calculate A F, considering triangle B F A

Using Pythagoras Theorem, we have:

$A\ B^2 = B\ F^2 + A\ F^2$

Substitute values for B F and A B;

$13^2 = 9^2 + A\ F^2$

$169= 81 + A\ F^2$

$A\ F^2 = 169 - 81$

$A\ F^2 = 88$

$A\ F = \sqrt{88$

$A\ F = 9.4$

Since F lies on line A C:

$A\ C = A\ F + F\ C$

$A\ C = 9.4 + 5.6$

$A\ C = 15.0$

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$Perimeter = A\ B + B\ C + A\ C$

$Perimeter = 13 + 10.6 + 15.0$

$Perimeter = 38.6$

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