<span>y = sin^-1(x/6), y = 0, and x = 6 so look at this
since x = 6 "(x/6)" would be (6/6)
Hope this helps and mark as brainliest please!</span>
Ask your teacher that’s all you have to do.
Completely different...
.15*.15*.15*500
500(.15^3)=1.6875
500(.45)=225
Each child gets =7×8=56crayoms
- One box contains 12 crayons
Total boxes
Round to next whole as we can't have depict
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.
