We know that m ║ n.
Let's first find the value 'x'.
When two lines are parallel, and a transversal is drawn, the angles on the same side of the transversal are equivalent.
This means that (5x + 16)° and (7x + 4)° are equivalent.
Equating them,
5x + 16 = 7x + 4
16 - 4 = 7x - 5x
12 = 2x
x = 12/2
x = 6°
Since we know the value of 'x', let's substitute them into the angles and find out the actual measurements.
5x + 16 = 5 × 6 + 16 = 30 + 16 = 46°.
7x + 4 = 7 × 6 + 4 = 42 + 4 = 46°.
Now let's find the value of 'y'.
If you observe carefully, (7x + 4)° and (y + 6)° form a linear pair.
This means that both those angles should add upto 180°.
Using that theory, the following equation can be framed:
(y + 6)°+ (7x + 4)° = 180°
Since we know the actual value of (7x + 4)°, let's substitute that value and move ahead.
(y + 6)° + 46° = 180°
y + 6 + 46° = 180
y + 52° = 180°
y = 180° - 52°
y = 128°
Therefore, the values of 'x' and 'y' are 46° and 128° respectively.
Hope it helps. :)
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:
a. -14,2/3
b. -28,3
Step-by-step explanation: