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3 years ago

Help please! Identify an equation in point slope form for the line perpendicular to Y =-4x + 8 that passes through (-2, 7)

Mathematics

1 answer:

777dan777 [17]3 years ago

3 0

Answer:

D. $y-7=\frac{1}{4}(x+2)$

Step-by-step explanation:

The point-slope form of a line is given by:

$y-y_1=m(x-x_1)$

The line given to us has equation; $y=-4x+8$ .

The slope of this line is -4. The of the line perpendicular to this line is the negative reciprocal of the slope of the given line.

Our slope of interest is therefore; $m=\frac{1}{4}$

Since the point goes through (-2,7), we have $x_1=-2,y_1=7$ .

We plug in the slope and the point into the point-slope formula to obain;

$y-7=\frac{1}{4}(x--2)$

The required equation is:

$y-7=\frac{1}{4}(x+2)$

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3 years ago

What is the coefficient of the x5y5-term in the binomial expansion of (2x – 3y)10? 10C5(2)5(3)5 10C5(2)5(–3)5 –10C5(2)5(–3)5 10C

butalik [34]

ANSWER

$10C_5(2)^{5}( - 3)^5$

EXPLANATION

The given binomial expansion is:

${(2x - 3y)}^{10}$

Compare this to

${(a + b)}^{n}$

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We want to find the coefficient of the term

${x}^{5} {y}^{5}$

This implies that, r=5.

The terms in the expansion can be obtained using

$T_{r+1}=nC_ra^{n-r}b^r$

We substitute the given values to obtain;

$T_{5+1}=10C_5(2x)^{10-5}( - 3y)^5$

$T_{6}=10C_5(2x)^{5}(3y)^5$

$T_{6}=10C_5(2)^{5}( - 3)^5 {x}^{5} {y}^{5}$

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