I’m not sure what you’re options are but an equation in Y=mx+b form for this would be y= -3/2x + 5
M is always the slope
B is always the y intercept
Answer:
1. $8.55
2. $4.29
3. $17.00
4. $10.47
5. $40.31
Step-by-step explanation:
Answer:
Step-by-step explanation:
All u need to do is graph the points so for -4,5 go 4 left then go up
Answer:
|F net| = 20.22 N
θ ≈ 19.8°
Step-by-step explanation:
F net = 15N i + 8cos(60°)N i + 8sin(60°)N j
= 15N i + 8×½N i + 8×√3/2N j
= 15N i + 4N i + 4√3N j
= 19N i + 4√3N j
|F net| = √(19²+(4√3)²) = √(361+48) = √409 ≈ 20.22N
tan(θ) = 4√3 ÷ 19 ≈ 0.36 → θ ≈ arctan(0.36) = 19.8°
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.
