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.
I hope this helps you
2x^2-4x+7=0 a=2 b=-4 c= 7
disctirminant =b^2-4ac
disctirminant =(-4)^2-4.2.7
disctirminant = 16-56= -40
x=-b+square root of disctirminant ÷2a
x=4+2square root of (-10)/4
x=2+square root of (-10)/2
x'=4 -2 square root of (-10)/4
x'=2 -square root of (-10)/2
Y=x i hope this helps you!
Answer:
Y= -4X + 8
Step-by-step explanation:
Use rise/run to find slope (y2- y1/ x2-x1). Plug the slope into point-slope form [y-y1= m(x-x1)]. this is your answer
Due to the fact that mode means most present number your answer would be mode or 100 dollars
