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.
34 Degrees because -7 - 27 = -34
4x - 24 = -2x + 30
4x + 2x = 30 + 24
6x = 54
x = 9
answer: the number is 9
Cross multiplying we get
x^2 = 2*7x
x^2 = 14x
x^2 - 14x = 0 is the answer
(a). To calculate Y at equilibrium
Y = C + I + G
Y = 40 + 0.8(Y – 0 + 10 + 20)
Y = 350
(b). To calculate C, I and G at Equilibrium
C = 40 + 0.8 Y
Since Y = 350
C = 40 + 0.8(350)
C = 40 + 280
C = 320
I = 20
G = 10
(c).To find
equilibrium Y
Given,
EX = 4 + 3EP/P
IM = 8 + 0.1 (Y - T) - 2EP/P
E = 3
P = 1
P = 1.5
Y = 170
