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2 years ago

Points A, B and C are collinear. Point B is between A and C. Find x if AC=20, BC=x-19 and AB=x+17.

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

4 0

Answer:

Step-by-step explanation:

20-[(x-19)+(x+17)]

=20-2x+2

=22-2x

As AB+BC=AC, 22-2x=0.

Thus 2x=22

x=11

it should be right

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Using the figure below identify 1 pair of the following angle relationships

tangare [24]

$\\1) \: \: \underline{ \sf{ Alternate \: interior \: angles \: \: : }}\\$

$\sf \implies \: 2\: \: and \: \: 7\\$

$\sf \implies \: 6\: \: and \: \: 3\\\\$

$\\2) \: \: \underline{ \sf{ Alternate \: interior \: angles \: \: : }}\\\\$

$\sf \implies \: 1\: \: and \: \: 8\\$

$\sf \implies \: 5\: \: and \: \: 4\\\\$

$\\3) \: \: \underline{ \sf{ Corresponding \: angles \: \: : }}\\$

$\sf \implies \: 3\: \: and \: \: 1\\$

$\sf \implies \: 5\: \: and \: \: 7\\$

$\sf \implies \: 4\: \: and \: \: 2\\$

$\sf \implies \: 6\: \: and \: \: 8\\\\$

$\\4) \: \: \underline{ \sf{ Consecutive\: Interior \: angles \: \: : }}\\$

$\sf \implies \: 2\: \: and \: \: 3\\$

$\sf \implies \: 6\: \: and \: \: 7\\\\$

$\\5) \: \: \underline{ \sf{ Consecutive\: Exterior \: angles \: \: : }}\\$

$\sf \implies \: 1\: \: and \: \: 4\\$

$\sf \implies \: 5\: \: and \: \: 8\\\\$

7 0

1 year ago

PLEASE HELP!!!! WILL MARK BRAINLIEST!!!!

irga5000 [103]

Answer:

None of these.

Step-by-step explanation:

Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.

Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.

In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.

If we look at the first term x^3 and we divide it by 13 we get $\frac{x^3}{13}$ we cannot reduce it so it is not a fraction so 13 is not a factor of P