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3 years ago

5

Steve is buying apples for the fifth grade.each bag holds 12 apples.if there are 75 students total,how many bags of apples will

Steve need to buy if he wants to give one apple to each student?

2 answers:

kirill115 [55]3 years ago

7 0

72 divided by 12 equals 6 apples each student

densk [106]3 years ago

7 0

7 bags he will have enough for each student to get 1 Apple and will also have 9 extra apples

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Marsha earn $25 per hour up to 36 hours; after this,she is paid double time. In one particular week she earned $1550.00 .how man

mel-nik [20]

Answer: Marta worked 13 hours overtime.

Step-by-step explanation:

1): You do 25 x 36 to get the amount of money she can make before she starts to make double.

36 x 25 = 900

2): Minus $900 from $1550

$1550 - $900 = $650

3): Divide $650 by $50

$650 / $50 = 13 hours worked overtime

7 0

3 years ago

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$

3.1.1.

$f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3$

Differentiate one term at a time:

• power rule

$\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4$

$\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}$

$\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}$

The last two terms are constant, so their derivatives are both zero.

So you end up with

$f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}$

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2 years ago

Find the integral of 3/sqrt of 1-4x^2

erastovalidia [21]

$\int {\frac{3}{\sqrt{1-4x^{2}}}} \, dx$
$= 3\int {\frac{1}{\sqrt{1-4x^{2}}}} \, dx$
$= 3\int {\frac{1}{\sqrt{4(\frac{1}{4} - x^{2})}}} \, dx$
$= \frac{3}{2}\int {\frac{1}{\sqrt{\frac{1}{4} - x^{2}}}}\, dx$

$= \frac{3}{2}sin^{-1}(2x) + C$

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antoniya [11.8K]

A

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2 years ago