Rs=6y+5,st=2y-3, and rt=12y-14 find y. r s t

Note that RT is the whole line segment, and RS and ST are parts of it

RS = 6y + 5

ST = 2y - 3

RT = 12y - 14

Set the equation

RS + ST = RT

(6y + 5) + (2y - 3) = 12y - 14

Simplify. Combine like terms

6y + 2y + 5 - 3 = 12y - 14

(6y + 2y) + (5 - 3) = 12y - 14

8y + 2 = 12y - 14

Note the equal sign. What you do to one side, you do to the other. Isolate the variable (y). Subtract 12y from both sides, and 2 from both sides.

8y (-12y) + 2 (-2) = 12y (-12y) - 14 (-2)

8y - 12y = -14 - 2

Simplify. Combine like terms

(8y - 12y) = (-14 - 2)

-4y = -16

Isolate the y. Divide -4 from both sides

-4y/-4 = -16/-4

y = (-16)/-4

y = 4

4 is your answer for y.

hope this helps

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.

146%

step-by-step explanation: