The answer is (-21, 13) for The second endpoint.
Let's start by calling the known endpoint L and the unknown K. We'll call the midpoint M. In order to find this, we must first note that to find a midpoint we need to take the average of the endpoints. To do this we add them together and then divide by 2. So, using that, we can write a formula and solve for each part of the k coordinates. We'll start with just x values.
(Kx + Lx)/2 = Mx
(Kx + 1)/2 = -10
Kx + 1 = -20
Kx = -21
And now we do the same thing for y values
(Ky + Ly)/2 = My
(Ky + 7)/2 = 10
Ky + 7 = 20
Ky = 13
This gives us the final point of (-21, 13)
Answer:
5
Step-by-step explanation:
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.
Answer:
C) 29x=5
reason:
first, distribute the parentheses
this will make your equation 8-20x=6-3+9x
simplify to get 8-20x=3+9x
add 20x to both sides to get 8=3+29x
subtract 3 from both sides to get 5=29x
Answer:
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