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3 years ago

8

Determine the whether the sum of 7/8 and 4/9 is closest to 0,1,2,or 3

1 answer:

elena55 [62]3 years ago

8 0

Whenever you add fractions it means you need a common denominator. The quickest way to find this number is to multiply the two denominators together (in this case 8 and 9).
So your new fractions are 63/72 and 32/72. When you add those you get 95/72.
Next simply divide the two numbers to get a decimal, so here we get: 1.31944...
So this number is closer to 1
Hope that helps!!

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Eva8 [605]

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3 years ago

Read 2 more answers

Differentiate the following functions (x^3+1)(x-2)÷x^2

Masja [62]

Given:

The given function is:

$\dfrac{(x^3+1)(x-2)}{x^2}$

To find:

The differentiation of the given function.

Solution:

Consider the given function,

$y=\dfrac{(x^3+1)(x-2)}{x^2}$

It can be written as:

$y=\dfrac{(x^3)(x)+(x^3)(-2)+(1)(x)+(1)(-2)}{x^2}$

$y=\dfrac{x^4-2x^3+x-2}{x^2}$

$y=\dfrac{x^4}{x^2}-\dfrac{2x^3}{x^2}+\dfrac{x}{x^2}-\dfrac{2}{x^2}$

$y=x^2-2x+\dfrac{1}{x}-\dfrac{2}{x^2}$

Differentiate with respect to x.

$y'=\dfrac{d}{dx}(x^2)-\dfrac{d}{dx}(2x)+\dfrac{d}{dx}(x^{-1})-\dfrac{d}{dx}\2(x^{-2})$

$y'=2x-2(1)+(-x^{-2})-2(-2x^{-3})$

$y'=2x-2-\dfrac{1}{x^2}+\dfrac{4}{x^3}$

Therefore, the differentiation of the given function is $2x-2-\dfrac{1}{x^2}+\dfrac{4}{x^3}$ .

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3 years ago