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BlackZzzverrR [31]
3 years ago
8

Find an equation of the line that is perpendicular to the line y=5x+6 and passes through the point (-2,5)

Mathematics
1 answer:
DaniilM [7]3 years ago
5 0

Answer:

5y=-x+23\\

Step-by-step explanation:

Since the equation of the line is perpendicular to the line y = 5x +6

we can use the formula:

m_1m_2=-1\\

where m1 is the slope of the line we have and,

m2 is the slope of the line we need

so here m1 is 5 as we compare it with the slope-intercept form our slope is:

y=5x+6\\y=mx+b

our m1 is 5 but we need m2 so plugging in the value we get m2 to be:

m_1m_2=-1\\5m_2=-1\\m_2=-1/5\\

and now we have our m2, now furthermore since the line passes through the point (-2 , 5) it means the point must lie on the line and satisfy it. We can use this point and the slope m2 and find out the equation by the point-slope form equation:

y-y_1=m(x-x_1)\\y-5=-\frac{1}{5}(x-(-2))\\\\5(y-5)=(-1)(x+2)\\5y-25=-x-2\\5y=-x-2+25\\5y=-x+23\\

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Step-by-step explanation:

I assume the digits cannot be repeated in the numbers.

and that we don't want any numbers that start with 0, as that would be considered a 3- digit number, right ?

so, under these assumptions we have the question of in how many ways can we pull 4 digits out of 6.

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we allow 3 digits, so we have

3×1/6 × 300 = 1/2 × 300 = 150

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