Determine whether equation is parallel or perpendicular or neither 9x+3y =12 and 12x +4y =7

1 answer:

5 0

The standard linear equation is y=mx+b
To be perpendicular the slope (m) has to be a negative.
reciprocal (2 and -1/2)
To be parallel the slope (m) has to be equal.
So put the equations in standard for
First you get:
3y=-9x+12 and
4y=-12x+7
You then have to divide the terms in front of the y's. You have to divide.
You get:
y=-3x+4 and
y=-3x+7/4
The -3's are the slopes (m)
-3=-3 so:
The lines are parallel

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indicate whether the measures 8, 14, and 16 can be the side lengths of a triangle. if they can, classify the triangle.

Aleksandr-060686 [28]

Answers:

A triangle is possible.

The triangle is acute and scalene

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Explanation:

We use the triangle inequality theorem to see if a triangle is possible with these side lengths.

Consider the side lengths a, b, and c. A triangle is possible if and only if the following 3 conditions hold true

a+b > c
a+c > b
b+c > a

Basically pick any two sides. If the sum is larger than the third side, then a triangle is possible.

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In this case we have: a = 8, b = 14, c = 16. The order of the a,b,c values doesn't really matter.

Then:

a+b = 8+14 = 22 which is larger than c = 16. So a+b > c is true.
a+c = 8+16 = 24 is larger than b = 14. So a+c > b is true.
b+c = 14+16 = 30 is larger than a = 8. So b+c > a is true.

All three inequalities mentioned are true; therefore, a triangle is possible with these side lengths.

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We'll use those a,b,c values in the converse of the pythagorean theorem to determine what kind of triangle this is. So far we know it's scalene because all three side lengths are different. It would be isosceles if it had exactly two equal sides, and equilateral if all 3 sides were the same length.

$a^2+b^2 = c^2\\\\8^2+14^2 = 16^2\\\\260 = 256\\\\$