as AB || DE then
angle BAE = angle AED
so angle AED = 70
as AD = AE then
angle AED = angle ADE = 70
now in a triangle
angle DAE + Angle ADE + angle AED = 180
angle DAE + 70 + 70 = 180
angle DAE = 180 - 140 = 40
5 is to 7 as is to 49 is to 51
10+6x=4x-4
6x-4x=-4-10
2x=-14
x-7
10=6(-7)=-32
4(-7)-4=-32
=-32
Answer:
choice 3) acute and equilateral
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.
