In the following diagram (used in Exercise #5), ∠ACD is known as an exterior angle of the triangle. ∠1 and ∠2 are known as remote interior angles relative to ∠ACD. Explain why m∠ACD = m∠1+ m∠2

why does it make sense to claim that an exterior angle to a triangle must always be greater in measure than either of its two remote interior angles?

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step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

in the math equation 8 x 7 = 56, the 8 and the 7 are both factors of 56, since in this lesson we will use square tiles to figure out the factors of a whole find as many different arrays as you can, but follow one important rule: all the rows must have the i can make two rows of nine, so 2 and 9 are both factors of 18.

step-by-step explanation:

nswer: a   y-axis

step-by-step explanation: