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goldfiish [28.3K]
3 years ago
6

Graph the line that passes through( -1,5)perpendicular to a line whose slope is -1/6

Mathematics
1 answer:
svetlana [45]3 years ago
4 0

Answer:

y - 5 = 6(x+1)

Step-by-step explanation:

if it is perpendicular to a line whose slope is -1/6, the slope of this line is the negative reciprocal which is 6.

y - 5 = 6(x+1)

this is point slope form btw

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Step-by-step explanation:

For step by step explanation see the pucture attached

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A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.
Crank

Answer: 1.25

Step-by-step explanation:

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Let Y be a random variable that denotes the scores in the exam.

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6 0
3 years ago
Inequality to x2 &lt; 64
Anarel [89]
              NOT MY WORDS TAKEN FROM A SOURCE!

(x^2) <64  => (x^2) -64 < 64-64 => (x^2) - 64 < 0 64= 8^2    so    (x^2) - (8^2) < 0  To solve the inequality we first find the roots (values of x that make (x^2) - (8^2) = 0 ) Note that if we can express (x^2) - (y^2) as (x-y)* (x+y)  You can work backwards and verify this is true. so let's set (x^2) - (8^2)  equal to zero to find the roots: (x^2) - (8^2) = 0   => (x-8)*(x+8) = 0       if x-8 = 0 => x=8      and if x+8 = 0 => x=-8 So x= +/-8 are the roots of x^2) - (8^2)Now you need to pick any x values less than -8 (the smaller root) , one x value between -8 and +8 (the two roots), and one x value greater than 8 (the greater root) and see if the sign is positive or negative. 1) Let's pick -10 (which is smaller than -8). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive
2) Let's pick 0 (which is greater than -8, larger than 8). If x=0, then (x^2) - (8^2) = 0-64 = -64 <0  so it is negative3) Let's pick +10 (which is greater than 10). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive Since we are interested in (x^2) - 64 < 0, then x should be between -8 and positive 8. So  -8<x<8 Note: If you choose any number outside this range for x, and square it it will be greater than 64 and so it is not valid.

Hope this helped!

:)
7 0
3 years ago
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