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Korolek [52]
3 years ago
9

Need help on 11 and 13 please

Mathematics
2 answers:
Arturiano [62]3 years ago
3 0

Answer:

11.       15x + 6

11. b       x + 8 - 16

13.          7x + 31

777dan777 [17]3 years ago
3 0

Answer:

11 A: x15 + 6

11 B: (x + 8) - 16

13: 7x + 31

Step-by-step explanation:

12x + 23 − 5x + 8

= 12x + 23 + −5x + 8

Combine Like Terms

= 12x + 23 + −5x + 8

= (12x + −5x) + (23 + 8)

= 7x + 31

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Pringles is planning their sales force spending for the next year. Each of their sales people spends about 3400 hours annually a
LUCKY_DIMON [66]

Answer:

459 sales people.

Step-by-step explanation

Time per call (t) = 45 min = 0.75 h

Hours per sales person (H) = 3,400 hours

Number of customers (n) = 40,000 customers

Call frequency (f)= 52 calls per year

The total number of sales people (S) needed, is given by the total time spent on calls for the year, divided by the amount of hours each person spends on sale:

S=\frac{40,000*0.75*52}{3,400}\\ S=458.8

Rounding up to the next whole person, Pringles needs 459 sales people.

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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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Valarie made fruit drinks for a birthday party. She used 3/6 of a cup of grape juice, 5/6 of a cup of strawberry juice, and 2/6
Nana76 [90]

Answer:

Valarie make 1\frac{2}{3}\ cups of fruit drink.

Step-by-step explanation:

Given:

Valarie made fruit drinks for a birthday party.

She used 3/6 of a cup of grape juice, 5/6 of a cup of strawberry juice, and 2/6 cup of soda. She mixes all the juices together.

Now, to find cups of fruit drink Valarie make.

As given that she mixes all juices altogether.

So, to get the cups of fruit drink we add all quantities of juice:

3/6 cup of grape juice + 5/6 cup of strawberry juice + 2/6 cup of soda.

=\frac{3}{6} + \frac{5}{6} +\frac{2}{6}

=\frac{3+5+2}{6}

=\frac{10}{6}

<em>Dividing numerator and denominator by 2 we get:</em>

=\frac{5}{3} =1\frac{2}{3}

Therefore, Valarie make 1\frac{2}{3}\ cups of fruit drink .

8 0
3 years ago
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