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:
n=2
Step-by-step explanation:
Answer:
x = 60
1) 60
2) 360
3) 160
60 + 360 + 160 = 580
Step-by-step explanation:
1) x
2) 6x
3) x + 100
x + 6x + x + 100 = 580
8x = 580 - 100
x = 480/8
x = 60
Answer:
b
Step-by-step explanation:
you can see in the figure, the function decrease from -1 to +3
5/16 = n/80
bring 80 over to the other side to leave n alone. 80 then gets a multiplication sign in front of it.
5/16 x 80=n
5/16=0.3125
0.3125x80=25
n=25
hope this helps
