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Wewaii [24]
2 years ago
12

Before it started to rain, Bobby cut 1 8 of the grass in his yard and Danny cut 1 5 of the grass in the yard. Both boys were dis

appointed because they were not able to cut the entire yard.
Which statement is TRUE?

A)Together the boys didn't even cut half the yard. B) The boys were able to cut more than half the yard. C) The boys cut almost three-fourths of the yard. D) Together the boys cut almost the entire yard.
Mathematics
jasmie
2 years ago
do you knw your math why are you talking little lady
2 answers:
Serggg [28]2 years ago
5 0

It’s B. The boys were able to cut more than half the yard.

blsea [12.9K]2 years ago
4 0

Answer:

A) Together the boys didn't even cut half the yard.

Step-by-step explanation:

The amount cut was ...

1/8 + 1/5 = 5/40 +8/40 = 13/40

This amount is less than half = 20/40, so choice A is appropriate.

_____

You can reason about these fractions several ways:

1. half is 2 1/2 fifths. 1/8 is less than 1/5, so 1/5+1/8 < (2 1/2)/5.

2. half is 4/8. 1/5 is more than 1/8 but is less than 2/8. 1/8 + 1/5 < 3/8 < 1/2

3. 1/5 < 1/4; 1/8 < 1/4; 1/2 = 2/4, so two amounts less than 1/4 cannot be more than 1/2.

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<u>Differentiate using the Quotient Rule</u> –

\qquad\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\

According to the given question, we have –

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Let's solve it!

\qquad\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2)  \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5)  -  ( x^3+5x+2) 2x}{(x^2-1)^2 }\\

\qquad\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

7 0
2 years ago
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