Answer:
343 yd
Step-by-step explanation:
Answer:
17
Step-by-step explanation:
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:
x
−
y
= −
11
Step-by-step explanation:
Answer:
(14-9i)+(2+3i) = 16-6i
Step-by-step explanation:
We need to find the equivalent of (14-9i)+(2+3i).
We need to add the two expressions.
(14-9i)+(2+3i) = 14+2 -9i+3i
= 16-6i
Here, the real part is 16
The imaginary part is -6
Hence, the required answer is equal to 16-6i.
