(5 + 4 - 2) x (-2)
= (9 - 2) x (-2)
= (7) x (-2)
= - 14 (Answer A)
f<span>(x)</span>=<span>x^2</span>−<span>6
</span>Replace <span>f<span>(x)</span></span> with <span>yy</span>.
<span>y=<span>x^2</span>−<span>6
</span></span>Interchange the variables.
<span>x=<span>y2</span>−6
</span>Solve for <span>yy</span><span>.
</span>
Move <span><span>−6</span><span>-6</span></span> to the right side of the equation by subtracting <span><span>−6</span><span>-6</span></span> from both sides of the equation.<span><span><span>y2</span>=6+x</span><span><span>y2</span>=6+x</span></span>Take the <span><span>square</span><span>square</span></span> root of both sides of the <span><span>equation</span><span>equation</span></span> to eliminate the exponent on the left side.<span><span>y=±<span>√<span>6+x</span></span></span><span>y=±<span>6+x
</span></span></span>The complete solution is the result of both the positive and negative portions of the solution.
Tap for more steps...
<span>y=<span>√<span>6+x</span></span>,−<span>√<span>6+x</span></span></span>
Solve for y<span> and replace with </span><span><span>f^<span>−1</span></span><span>(x).
</span></span>
<span>Answer is f<span>−1</span></span><span>(x)</span>=<span>√<span>6+x</span></span>,−<span>√<span>6+<span>x</span></span></span>
JKM = M + L
15x - 48 = 40 + 5x + 12
10x - 48 = 52
10x = 100
x = 10
JKM = 15(10) - 48 = 102
180 = JKM + MKL
180 = 102 + MKL
MKL = 180 - 102 = 78
As I've shown on picture that I attached:
∠ADC=∠ADB+∠BDC⇒134=5x+4+7x-2
12x+2=134
12x=132⇒x=132/12
x=11
Angle BDC measure 7x-2=7*11-2⇒∠BDC=75.
