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.
Answer:
option B
Given : |x + 4| < 5
A. –5 > x + 4 < 5
B. –5 < x + 4 < 5
C. x + 4 < 5 and x + 4 < –5
D. x + 4 < 5 or x + 4 < –5
In general , |x|< n where n is positive
Then we translate to -n < x < n
|x + 4| < 5
5 is positive, so we translate the given absolute inequality to
-5 < x+4 < 5
So option B is correct
Answer:
you would need to see which one adds up or just an answer out in the wen then see what you get from there if not go find another way
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>--------------------------------</u>
hope it helps...
have a great day!!!
