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:
the answer I believe is c
Step-by-step explanation:
Step-by-step explanation:
It is (A)95° by using External Angle Property of triangles.
BTW whats up with that order C-D-B-A
X is left and right
hmm
see the aproximate slope
see endpoints
ok, so from about (0,3) to (35,30)
slpoe is 27/35
so
y=(27/35)x+3 about
so when x=
ok, so according to the garph, when x=35, y=30, not 20
ok, so when x=40
y=(27/35)x+3
y=(27/35)*40+3
y=33.8571 about for x=40