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3 years ago

I am giving brainlist to quickest and the correct answer

Mathematics

2 answers:

Natalka [10]3 years ago

5 0

Answer:

Step-by-step explanation:

Sally drove 300 miles in 6 hours, so all you have to do is divide 300 by 6.

300/6=50

Sally drove 50 miles per hour.

Have an amazing day...I hope this helped!! c:

kirza4 [7]3 years ago

4 0

50 miles drive per hour

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Can i get help on 18 & 22 please

Pavlova-9 [17]

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3 years ago

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Savatey [412]

The slope of the line is -3/4

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3 years ago

Write the fraction equovalent of 0.82 in simplest form. use the / bottom as the fraction bar.

Luden [163]

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3 years ago

Malory isn’t putting money in two savings accounts.Account A started with $150 and Account B started with $300.Malory deposits $

EastWind [94]

Given:

There are two savings accounts - account A has $150 and account B has $300 to begin with.
Malorie deposits $16 in account A and $12 in account B every month.

To Find:

The number of months where the balance of both is equal and the balance itself.

Solution:

In 37.5 months, the account balances in both accounts will be equal.

This balance will be equal to $750 in both account A and B.

Calculation:

Let x be the number of months where both the accounts have equal balance.

So, at the end of x months, 16x is the amount of money that will be added to account A (since $16 is added every month).

And the total amount at the end of x months will therefore be $(150+16x)

Similarly, at the end of x months, 12x is the amount of money that will be added to account B (since $12 will be added every month).

And the total amount at the end of x months will therefore be $(300+12x)

By our assumption, x is the number of months it will take for both the accont balances will be equal. Therefore, we have the equation

$150+16x=300+12x\\4x=150\\x=37.5$

Therefore, in 37.5 months, the account balances in both accounts will be equal.

This balance will be equal to $150+16(37.5)=750$ i.e. $750 will be the balance in both account A and B.

8 0

3 years ago

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

<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 –

f(x) = x^3+5x+2
g(x) = x^2-1

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