Answer:
the aswer is 2
Step-by-step explanation:
this is because 10/5=2
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.
3x-4y+2 hahahaha its not me pep perillp ie eewewr its jwghvjbe rijwte ru
Answer:
9
Step-by-step explanation:
We know that for 
where p and q are different prime numbers the number of positive divisors are (m+1) (n+!)
We have given 
So here m=2 and n=2
So number of positive divisors 
=9
So the number of positive divisors of is 9
When adding fractions you need a common denominator so lets look at what "fractions" we are dealing with:2/3 1/6 and 1/4.
