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3 years ago

5

Factor 5(x^2n-1)·(2x^3n 1)^2

1 answer:

nikdorinn [45]3 years ago

3 0

I think I may be wrong check

5(x^2n-1)×(2x^3n-1) ^2

=20n3 x8 -20n2 x5 +20n x3 +20n x3 +5n x2-5

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2 years ago

A line passes through the point (-8, 7) and has a slope of negative 3/4Write an equation in slope-intercept form for this line.

Arlecino [84]

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Step-by-step explanation:

We do not have enough information for slope intercept form. But we can use the point-slope formula to find the information. The formula is $y -y_{1} =m(x -x_{1})$ where we substitute a point (x,y) for $(x_{1},y_{1})$ .

We have m=-3/4 and (-8, 7). We input m and $x_{1} =-8\\y_{1}=7$ .

$y-7=-\frac{3}{4} (x-(-8))\\y-7=-\frac{3}{4} (x+8)$

We now simplify the parenthesis and solve for y.

$y-7=-\frac{3}{4} (x+8)\\y-7=-\frac{3}{4}x+-\frac{3}{4} (8)$

We convert 8 into a fraction with 1 as the denoinator.

$y-7=-\frac{3}{4}x+-\frac{3}{4} (\frac{8}{1} )\\y-7=-\frac{3}{4}x+-\frac{24}{4}\\y-7=-\frac{3}{4}x+-6$

We add 7 to both sides to isolate y,

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This is slope intercept form. The line as slope -3/4 and y-intercept (0,1) or b=1.

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3 years ago