Answer:
u can maybe help it depends on the questions
Answer:
1/1
Step-by-step explanation:
cause
1
Simplify \frac{1}{2}\imath n(x+3)21ın(x+3) to \frac{\imath n(x+3)}{2}2ın(x+3)
\frac{\imath n(x+3)}{2}-\imath nx=02ın(x+3)−ınx=0
2
Add \imath nxınx to both sides
\frac{\imath n(x+3)}{2}=\imath nx2ın(x+3)=ınx
3
Multiply both sides by 22
\imath n(x+3)=\imath nx\times 2ın(x+3)=ınx×2
4
Regroup terms
\imath n(x+3)=nx\times 2\imathın(x+3)=nx×2ı
5
Cancel \imathı on both sides
n(x+3)=nx\times 2n(x+3)=nx×2
6
Divide both sides by nn
x+3=\frac{nx\times 2}{n}x+3=nnx×2
7
Subtract 33 from both sides
x=\frac{nx\times 2}{n}-3x=nnx×2−3
Wish i could help you don’t have a lot of information
Answer:
Let f(x) = x^3 -6x^2 +11x-8+K
by factor theorem, f(3)=0
27-54+33-8+K=0
K=2
hope that helps?
Step-by-step explanation:
