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kupik [55]
4 years ago
8

What is the difference?

Mathematics
1 answer:
Crazy boy [7]4 years ago
6 0

Answer:

<h2>D. \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\ or StartFraction x squared + 3 x minus 12 Over (x + 3) (x minus 5) (x + 7) EndFraction</h2>

Step-by-step explanation:

Given the expression \frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }, the dfference is expressed as follows;

Step1: First we need to factorize the denominator of each function.

\frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }\\= \frac{x}{x^{2}-5x+3x-15 } - \frac{4}{x^{2} + 7x-5x - 35 }\\= \frac{x}{x(x-5)+3(x-5) } - \frac{4}{x( x+ 7)x-5(x +7) }\\= \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\\\

Step 2: We will find the LCM of the resulting expression

=  \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\= \frac{x(x+7)-4(x+3)}{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+7x-4x-12 }{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\

The final expression gives the difference

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The expression into a single logarithm is log[(x)^{10}][(2)^{30}]

Step-by-step explanation:

Let us revise some logarithmic rules

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  • nlog(a)+mlog(b)=log[(a)^{n}][(b)^{m}]

∵ 10 log(x) + 5 log(64)

- At first re-write 10 log(x)

∴  10 log(x) = log(x)^{10}

- Then re-write 5 log(64)

∴  5 log(64) = log(64)^{5}

∴ 10 log(x) + 5 log(64) = log(x)^{10} + log(64)^{5}

- Use the 3rd rule above to make it single logarithm

∵ log(x)^{10} + log(64)^{5} = log[(x)^{10}][(64)^{5}]

∴ 10 log(x) + 5 log(64) = log[(x)^{10}][(64)^{5}]

∵ 64 = 2 × 2 × 2 × 2 × 2 × 2

∴ We can write 64 as 2^{6}

∴ (64)^{5}=(2^{6})^{5}

- Multiply the two powers of 2

∴ (64)^{5}=(2)^{30}

∴ 10 log(x) + 5 log(64) = log[(x)^{10}][(2)^{30}]

The expression into a single logarithm is log[(x)^{10}][(2)^{30}]

Learn more:

You can learn more about the logarithmic functions in brainly.com/question/11921476

#LearnwithBrainly

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The total volume is (110 +80) yd^3 = 190 yd^3.
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