y = x^2 + 2x...eqn 1
y = 3x + 20...eqn 2
subst for y in eqn 1...
=> x^2 +2x = 3x +20
=> x^2 - x - 20 =0
=> (x-5) (x+4) =0
=> x = 5 or -4
for x =5, y = 35 (sub for x in eqn 1 or 2)
for x = -4, y = 8 (sub for x in eqn 1 or 2)
Devisor is the bottom number
hmm
11.75=11 and 3/4
so times it by 4 to clear denomenator
but do it by top and bottom
23.4/11.75 times 4/4=93.6/47
so times it by 4/4
Answer:
Step-by-step explanation:
![6k^3+10k^2-56k\\2k(3k^2+5k-28)\\=2k[3k^2+12k-7k-28]\\=2k[3k(k+4)-7(k+4)]\\=2k(k+4)(3k-7)](https://tex.z-dn.net/?f=6k%5E3%2B10k%5E2-56k%5C%5C2k%283k%5E2%2B5k-28%29%5C%5C%3D2k%5B3k%5E2%2B12k-7k-28%5D%5C%5C%3D2k%5B3k%28k%2B4%29-7%28k%2B4%29%5D%5C%5C%3D2k%28k%2B4%29%283k-7%29)
common factor is 3k-7
Larger because I want the points and you are going to get an F
