Answer:
2.125
Step-by-step explanation:
Answer:
18245
Step-by-step explanation:
We have to use L.C.M,
L.C.M(20,24,32,38)
2|20,24,32,38
2| 10 ,12 ,16 ,19
2| 5 , 6 , 8 , 19
2| 5 , 3 , 4 , 19
2| 5 , 3 , 2 , 19
L.C.M = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 x 19
= 18240
Now for each case remainder is 5,
So the number is 18240+5
=> 18245
Answer: 30 servers
5+2 = 7
42/7 = 6 sets of employees
5*6 = 30 servers
Step-by-step explanation:
5+2 = 7
42/7 = 6 sets of employees
5*6 = 30 servers
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 value of x would be 150
