$\dfrac{8^{3x}}{2^{10}}=\dfrac{4^{2x}}{16} \\\dfrac{2^{3(3x)}}{2^{10}}=\dfrac{4^{2x}}{4^2} \\2^{9x-10}=4^{2x-2} \\2^{9x-10}=2^{2(2x-2)} \\2^{9x-10}=2^{4x-4} \\9x-10=4x-4 \\5x-6=0 \\5x=6 \\\boxed{x=\dfrac{6}{5}}$

Hope this helps.

r3t40

3 0

3 years ago

Please answer and explain

sashaice [31]

8=2*2*2=2^3
32=2*2*2*2*2=2^5
[8^(2/3)] / [32^(2/5)] = [(2^3)^(2/3)] / [(2^5)^(2/5)] =
= [2^(3*2/3)] / [2^(5*2/5)]= [2^2] / [2^2] = 4/4 = 1
Answer: 1

4 0

2 years ago

If sin theta = - squrt.3/2 and pi < theta < 3pi/2, what are the values of cos theta and tan theta?

Kryger [21]

So this first wants you to find where sin is √3/2 when θ is between π and 3π/2. θ would therefore be located at 2π/3.

Now plug in the value of θ for cosine:

cos (2π/3) = -1/2

And tangent:

tan (2π/3) = -√3/3

7 0

2 years ago

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