Can anyone with 13 14 and 15 anything !

13. The midpoint divides the segment into two equal parts. That is

BC = (1/2)·AC = (1/2)·(8x -20)

BC = 4x -10

Since you are told the other half is 3x -1, the two halves being equal tells you

... AB = BC

... 3x -1 = 4x -10

... x = 9 . . . . . add 10-3x

Then the measures of AB and BC are 3·9 - 1 = 4·9 -10 = 26.

BC = 26

The problem statement is not clear as to the desired form of the answer—whether it is to be in terms of x, or a number.

14. Same deal as the previous problem. The halves of the bisected segment are equal, giving you an equation that can be solved for x.

... CG = GD

... 5x -1 = 7x -13

... 12 = 2x . . . . . . . . . add 13-5x

... x = 6 . . . . . . . . . . divide by 2

Now you have a value of x that you can use to find the length of EF.

... EF = 6x - 4 = 6·6 - 4 = 32

The parts of EF add to give its total length, so

... EG + GF = EF

... EG + 13 = 32

... EG = 19

15. Once again, the parts of a line segment add to give the whole.

... RS + ST = RT

... (2x -4) + (4x -1) = (8x -43)

... 38 = 2x . . . . . . . . . . . . . . . . add 43-6x

... x = 19

Since R is the midpoint of QS, RS will be half the length of QS. The length of RS is 2x -4 = 2(19) - 4 = 34. Then the length of QS is 2·34 = 68.

QS = 68

To help check our work, we can find the lengths of the other segments in the diagram. ST = 4x - 1 = 75; RT = 8x - 43 = 109. So, we have ...

... QR : RS : ST = 34 : 34 : 75 . . . with QS = 68 and RT = 109.

the anwser is d) 75°

step-by-step explanation: