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3 years ago

7

Kyle drives 330 miles on 15 gallons of gas. How many miles per gallon (mpg) did Kyle get? A) 12 mpg B) 16 mpg C) 19 mpg D) 22 mp

g

2 answers:

Vikki [24]3 years ago

5 0

Answer:

d. 22mpg

Step-by-step explanation:

330/22

mel-nik [20]3 years ago

5 0

22 becaUSE 330 per fifteen

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If 30 marks is 40% what is full marks

Ray Of Light [21]

The full marks is 75.

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3 years ago

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What are three consecutive odd integers whose product of the first and the third is for less than the square of the second

Inessa05 [86]

Answer:

in steps

Step-by-step explanation:

Are you trying to prove any 3 positive consecutive odd integers whose product of the first and the third is four less than the square of the second?

3 positive consecutive odd integers: x-2, x, x+2

(x-2) (x+2)=x² - 4 ..... correct

11, 13, 15 11 x 15 = 165 = 13² - 4

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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

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3 years ago

3 purple blocks, 3 yellow blocks, 6 green blocks, and 4 red blocks. Her favorite game is to stack all of the blocks one on top o

KatRina [158]

I believe the answer if there are 216 ways to do this, but I’m not entirely sure the question.

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4 years ago

Danielle poured 3/4 gallon of water from a 7/8 gallon bucket. How much water is left in the bucket?

Radda [10]

3/4 = 6/8 after you multiply top and bottom by 2

Subtract 3/4 from 7/8 and we get
7/8 - 3/4 = 7/8 - 6/8 = (7-6)/8 = 1/8

There is 1/8 of a gallon left

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3 years ago