20 and 10, 20 is 2 times 10 and if you add them together its 30
Unless K is defined, this problem doesn’t make sense...even if we divide over k then k would eventually equal nothing (not zero)
BC= DF
B(0,2) , C(3,2)
d= (3-0)^2 + (2-2)^2
d= 9
Square root 9= 3
DF = 3
Hope this helps!
Answer:
I cant see the whole thing
let's multiply both sides by the LCD of all fractions, in this case is 4, just to do away with the denominators for a few seconds
![\bf f(x) = \cfrac{1}{4}x^2+\cfrac{1}{2}x-\cfrac{35}{4}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4[f(x)]=4\left( \cfrac{1}{4}x^2+\cfrac{1}{2}x-\cfrac{35}{4} \right)} \\\\\\ 4f(x) = x^2+2x-35\implies 4f(x) = (x+7)(x-5) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill f(x) = \cfrac{1}{4}(x+7)(x-5)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%20%3D%20%5Ccfrac%7B1%7D%7B4%7Dx%5E2%2B%5Ccfrac%7B1%7D%7B2%7Dx-%5Ccfrac%7B35%7D%7B4%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B4%7D%7D%7B4%5Bf%28x%29%5D%3D4%5Cleft%28%20%5Ccfrac%7B1%7D%7B4%7Dx%5E2%2B%5Ccfrac%7B1%7D%7B2%7Dx-%5Ccfrac%7B35%7D%7B4%7D%20%5Cright%29%7D%20%5C%5C%5C%5C%5C%5C%204f%28x%29%20%3D%20x%5E2%2B2x-35%5Cimplies%204f%28x%29%20%3D%20%28x%2B7%29%28x-5%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20f%28x%29%20%3D%20%5Ccfrac%7B1%7D%7B4%7D%28x%2B7%29%28x-5%29~%5Chfill)