3 years ago

Please help, if able! 10 points up^^

Mathematics

1 answer:

Debora [2.8K]3 years ago

7 0

$cosine=\dfrac{adjacent}{hypotenuse}\\\\sine=\dfrac{opposit}{hypotenuse}\\\\\boxed{\cos57^o=\dfrac{x}{51}}$

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Simplify the expression. 8x^-10 y^'6 -2x^2y^-8 Write your answer without negative exponents.

sergij07 [2.7K]

Answer:

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^{14} - 2x^{12}}{x^{10}y^8}$

Step-by-step explanation:

Given

$8x^{-10}y^6 - 2x^2y^{-8}$

Required

Simplify

Rewrite as:

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6}{x^{10}} - \frac{2x^2}{y^8}$

Take LCM

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6*y^8 - 2x^2 * x^{10}}{x^{10}y^8}$

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In traveling from a dormitory to a certain city, a student went 1/5 of the way by foot, 2/3 of the way by bus, and the remaining

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Step-by-step explanation:

Now, what we know is that the total distance from the dormitory to the city is 1 if expressed in fraction. What we need to know is the fraction of the journey that is scheduled as the last part.

We can get this by subtracting the fractions of the first two phases from 1.

This goes as follows:

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Now, we know that the final 8 kilometers constitute a fraction of just 2/5

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