Use a proof by contradiction to prove that the sum of two odd integers is even cm

Sum of two odd integers is always even.

Step-by-step explanation:

Let m and n be two odd integers.

Since m and n are odd they can be written in the form m =2r + 1 and n = 2s + 1, where r and s are integers.

Let us suppose that their sum is not even.

m + n = (2r+1) + (2s + 1)

          = 2r + 2s + 2

          = 2(r+s+1)

          = 2z

Thus, the sum of m and n can be written in the form 2z where z is an integer. But this is a contradiction to the fact that their sum is even.

Hence, our assumption was wrong and the sum of two odd integers is always even.

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4

the answer is 11.2 (i know i’m late)