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.
The volume of cube is 1000 cubic feet
<em><u>Solution:</u></em>
Given that,
<em><u>The volume of cube is given by formula:</u></em>
Where,
V is the volume of cube
"s" is the length of one side
<em><u>What is the volume of a cube if the length of a side s is 10 feet</u></em>
Given, s = 10 feet
<em><u>Substitute s = 10 in given formula,</u></em>
Thus volume of cube is 1000 cubic feet
If all were children, revenue would be 700*$7 = $4900. Revenue is actually $6400 -4900 = $1500 more than that. Each adult admission that replaces a child's admission adds $10 -7 = $3 to the revenue, so there must have been
$1500/$3 = 500 . . . . adult admissions
There were 200 children at the show.
There were 500 adults at ths show.
