Answer:
Step-by-step explanation:
I think the one you want is the last one, but I've never called it that.
21x^2 + 147x + x^3 + 343
x^3 + 343 is a cubic binomial. It factors into (x + 7) (x^2 - 7x + 49)
21x(x + 7) + (x + 7)(x^2 - 7x + 49)
(x + 7) [ 21x - 7x + x^2 + 49)
(x + 7) [x^2 + 14x + 49]
(x + 7) ( x+7)^2
(x + 7)^3
It's the only answer that does factor. There must be a simpler way of doing it, but this will give you the right answer.
I can’t really help you too much with the graphing but what you do with ordered pairs is whatever is the first number (example for point A it’s -2) you go and find it on the horizontal line. Once you find it on the horizontal line you either go up or down the graph by how many units the second number is (4 so you would go up four in the graph) and then you plot the point
Answer:A
Step-by-step explanation:
Answer:
no bro that's a bad idea
Step-by-step explanation:
Answer:
Step-by-step explanation:
Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?
Let p(x)=kpx+dp and q(x)=kqx+dq than
f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7
p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7
(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)
So you want:
−2kpk2q=0
and
kpkq=−1
and
2kpd2p−3kpdq+7=0
Now I amfraid this doesn’t work as −2kpk2q=0 that either kp or kq is zero but than their product can’t be anything but 0 not −1 .
Answer: there are no such linear functions.
