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4 years ago

7

A pharmacist put 4.536 ounces of vitamin pills into bottles. She put 0.042 ounce of vitamin pills into each bottle. How many bot

tles did the pharmacist use for these vitamin pills?

1 answer:

Romashka [77]4 years ago

3 0

THE ANSWER IS 5 since the decimals are out of place

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9/6 × 4/5 <br> 1 <br>) 36/11<br>) 6/5<br>) 32/30<br>) 13/15

erik [133]

Answer:

6/5

Step-by-step explanation:

9×4=36

6×5=30

36/30÷2

18/15÷3

6/5

6 0

2 years ago

Simplify 2(3×8-15÷16)<br>

KengaRu [80]

Answer:

46.125 or 46 1/8

Step-by-step explanation:

2(3×8-15÷16)

=2(24-15÷16)

= 48 - 30/16

= (768 - 30)/16

= 738/16

= 46.125 or 46 1/8

8 0

4 years ago

What is the slope of (4,8) (2,4)

weqwewe [10]

2
$\frac{4 - 8}{2 - 4} = \frac{ - 4}{ - 2} = 2$

6 0

4 years ago

\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

Vinvika [58]

$\displaystyle \lim_{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)$

Both the numerator and denominator approach 0, so this is a candidate for applying L'Hopital's rule. Doing so gives

$\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)$

This again gives an indeterminate form 0/0, but no need to use L'Hopital's rule again just yet. Split up the limit as

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}$

Now recall two well-known limits:

$\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax}=1\text{ if }a\neq0 \\\\ \lim_{x\to0}\frac{\ln(1+ax)}{ax}=1\text{ if }a\neq0$

Compute each remaining limit:

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{\ln(1+3x)}{3x} \times \lim_{x\to0}\frac{3x}{\sin(3x)} = \frac23$

$\displaystyle \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\frac{1}{1+3x} = \frac23$

$\displaystyle \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\cos(x)}{\cos(3x)} = \frac13$

$\displaystyle \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\sin(x)}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\sec^2(x) = \frac13$

$\displaystyle \lim_{x\to0}\frac{6x^2}{3\sin(3x)} = \frac23 \times \lim_{x\to0}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}x = 0$

So, the original limit has a value of

2/3 + 2/3 + 1/3 + 1/3 - 0 = 2

6 0

3 years ago

Randy ate 29 crackers in 4 minutes. At what RATE did Randy eat?

lisabon 2012 [21]

Answer:

7 1/4 cookies per minute

Step-by-step explanation:

29/4 = 7 1/4

6 0

3 years ago

Read 2 more answers