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3 years ago

9

Line k passes through points (-3,-6) and (-1,12) what is the slope of the line perpendicular to line c

1 answer:

yan [13]3 years ago

4 0

Do you mean "perpendicular to k"? We have no line c there so I'm going with line k. First we need the slope of line k. Fitting those values into the slope formula we have $m= \frac{12-(-6)}{-1-(-3)} = \frac{18}{2} =9$ . So the slope of k is 9. The slope of a line perpendicular to k has the opposite sign and the fraction is flipped. The perpendicular slope then is -1/9.

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3 years ago

Which expression is equivalent to (5g^4+5g^3-17g^2+6g)-(3g^4+6g^3-7g^2-12)/g+2

ankoles [38]

Option C:

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Given expression:

$$\frac{\left(5 g^{4}+5 g^{3}-17 g^{2}+6 g\right)-\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right)}{g+2}$

To find which expression is equal to the given expression.

$$\frac{\left(5 g^{4}+5 g^{3}-17 g^{2}+6 g\right)-\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right)}{g+2}$

Expand the term $-\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right):-3 g^{4}-6 g^{3}+7 g^{2}+12$

$$=\frac{5 g^{4}+5 g^{3}-17 g^{2}+6 g- 3 g^{4}-6 g^{3}+7 g^{2}+12}{g+2}$

Arrange the like terms together.

$$=\frac{5 g^{4}- 3 g^{4}+5 g^{3}-6 g^{3}-17 g^{2}+7 g^{2}+6 g+12}{g+2}$

$$=\frac{2 g^{4}- g^{3}-10 g^{2}+6 g+12}{g+2}$

Factor the numerator $2 g^{4}-g^{3}-10 g^{2}+6 g+12=(g+2)\left(2 g^{3}-5 g^{2}+6\right)$

$$=\frac{(g+2)\left(2 g^{3}-5 g^{2}+6\right)}{g+2}$

Cancel the common factor g + 2, we get

$=2 g^{3}-5 g^{2}+6$

Hence option C is the correct answer.

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3 years ago

Read 2 more answers

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morpeh [17]

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