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3 years ago

Find the distance between parallel lines whose equations are y = x - 6 and y = x + 8.

Mathematics

1 answer:

Sidana [21]3 years ago

8 0

Answer:

Step-by-step explanation:

Distance between y2 = x + 8 and y1 = x - 6

y2 - y1 = (x + 8) - (x - 6) = 14

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3 years ago

Enter the correct answer in the box. IF x≠0, what is the sum of 4 v^3 x^10 + 5x^3 ^3

dsp73

Answer: $14*\sqrt[3]{x^{10}}$

Step-by-step explanation:

First, let's remember the rules:

$x^{1/n} = \sqrt[n]{x}$

such that:

$\sqrt[n]{x^m} = x^{m/n}$

$(x^a)^b = x^{a*b}$

$x^a*x^b = x^{a + b}$

Now we have the equation:

$4*\sqrt[3]{x^{10}} + 5*x^3*\sqrt[3]{8*x}$

First, let's rewrite the roots as we saw above:

$4*x^{10/3} + 5*x^3*\sqrt[3]{8}*x^{1/3} = 4*x^{10/3} + 5*\sqrt[3]{8}*x^{3 + 1/3}$

We also know that: $\sqrt[3]{8} = 2$

and that: $3 + 1/3 = 9/3 + 1/3 = 10/3$

Replacing those two things in our equation, we get:

$4*x^{10/3} + 5*\sqrt[3]{8}*x^{3 + 1/3} = 4*x^{10/3} + 5*2*x^{10/3} = (4 + 5*2)*x^{10/3}$

$(4 + 10)*x^{10/3} = 14*\sqrt[3]{x^{10}}$

Where in the final step, I returned to the cubic root form.

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3 years ago