Logarithm
1 answer:
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<h2>(1)</h2><h2> =(a+b)(3c-d)</h2><h2> =a(3c-d)+b(3c-d)</h2><h2> =3ac-ad+3bc-bd</h2><h2>(2)</h2><h2> =(a-b)(c+2d)</h2><h2> =a(c+2d)-b(c+2d)</h2><h2> =ac+2ad-bc-2bd</h2><h2>(3)</h2><h2> =(a-b)(c-2d)</h2><h2> =a(c-2d)-b(c-2d)</h2><h2> =ac-2ad-bc+2bd</h2><h2>(4)</h2><h2> =(2a+b)(c-3d)</h2><h2> =2a(c-3d)+b(c-3d)</h2><h2> =2ac-6ad+bc-3bd</h2>
Answer:
x = -2
Step-by-step explanation:
The tangent line has length "x + 8"
The secant line has length "x + 6 + 5", where
5 is the inner part
x + 6 is the outer part
Now,
the secant-tangent theorem tells us that square of the tangent line is equal to the outer segment of secant line multiplied by length of whole secant line.
So, we can say:
We can solve for x shown below:
The value of x is -2