well, we know it's a rectangle, so that means the sides JK = IL and JI = KL, so
![\stackrel{JK}{3x+21}~~ = ~~\stackrel{IL}{6y}\implies 3(x+7)=6y\implies x+7=\cfrac{6y}{3} \\\\\\ x+7=2y\implies \boxed{x=2y-7} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{JI}{6y-6}~~ = ~~\stackrel{KL}{2x+20}\implies 6(y-1)=2(x+10)\implies \cfrac{6(y-1)}{2}=x+10 \\\\\\ 3(y-1)=x+10\implies 3y-3=x+10\implies \stackrel{\textit{substituting from the 1st equation}}{3y-3=(2y-7)+10} \\\\\\ 3y-3=2y+3\implies y-3=3\implies \blacksquare~~ y=6 ~~\blacksquare ~\hfill \blacksquare~~ \stackrel{2(6)~~ - ~~7}{x=5} ~~\blacksquare](https://tex.z-dn.net/?f=%5Cstackrel%7BJK%7D%7B3x%2B21%7D~~%20%3D%20~~%5Cstackrel%7BIL%7D%7B6y%7D%5Cimplies%203%28x%2B7%29%3D6y%5Cimplies%20x%2B7%3D%5Ccfrac%7B6y%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20x%2B7%3D2y%5Cimplies%20%5Cboxed%7Bx%3D2y-7%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7BJI%7D%7B6y-6%7D~~%20%3D%20~~%5Cstackrel%7BKL%7D%7B2x%2B20%7D%5Cimplies%206%28y-1%29%3D2%28x%2B10%29%5Cimplies%20%5Ccfrac%7B6%28y-1%29%7D%7B2%7D%3Dx%2B10%20%5C%5C%5C%5C%5C%5C%203%28y-1%29%3Dx%2B10%5Cimplies%203y-3%3Dx%2B10%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20from%20the%201st%20equation%7D%7D%7B3y-3%3D%282y-7%29%2B10%7D%20%5C%5C%5C%5C%5C%5C%203y-3%3D2y%2B3%5Cimplies%20y-3%3D3%5Cimplies%20%5Cblacksquare~~%20y%3D6%20~~%5Cblacksquare%20~%5Chfill%20%5Cblacksquare~~%20%5Cstackrel%7B2%286%29~~%20-%20~~7%7D%7Bx%3D5%7D%20~~%5Cblacksquare)
Answer:
14
Step-by-step explanation:
In a triangle, the midline joining the midpoints of two sides is parallel to the third side and half as long, so
FB = EC/2
2x - 6 = 44/2
2x - 6 = 22
2x = 22 + 6
2x = 28
x = 28/2
x = 14
Idk but just divide both of the fraction hope i helped you a little
Answer: See explanation
Step-by-step explanation:
You didn't give the expressions but here are some expressions
a. ✓4x²y^4. 1. 2x✓y
b. ✓8x²y. 2. 2y✓2x
c. ✓4x²y. 3. 2xy²
d. ✓16xy². 4. 2x✓2y
e. ✓8xy². 5. 4y✓x
a. ✓4x²y^4 = ✓4 × ✓x² × ✓y^4
= 2 × x × y²
= 2xy²
Therefore, ✓4x²y^4 = 2xy²
b. ✓8x²y = ✓8 × ✓x² × ✓y
= ✓4 × ✓2 × ✓x² × ✓y
= 2 × ✓2 × x × ✓y
= 2x✓2y
Therefore, ✓8x²y = 2x✓2y
c. ✓4x²y = ✓4 × ✓x² × ✓y
= 2 × x × ✓y
= 2x✓y
Therefore, ✓4x²y = 2x✓y
d. ✓16xy² = ✓16 × ✓x × ✓y²
= 4 × ✓x × y
= 4y✓x
Therefore, ✓16xy² = 4y✓x
e. ✓8xy² = ✓8 × ✓x × ✓y²
= ✓4 × ✓2 × ✓x × ✓y²
