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2 years ago

15

Juan runs out of gas in a city. He walks 30 yards west and then 16yards south looking for a gas station. How far is he from his

starting point?

1 answer:

iragen [17]2 years ago

5 0

We need to calculate the hypotenuse; Juan is located on a right triangle.
34 yards

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What is 97 divided by 7

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7 can go into 97 13 times w/ a remainder of 6.

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2 years ago

A chef plans to mix 100% vinegar with Italian dressing. The Italian dressing contains 12%

Hitman42 [59]

Answer:

Amount of vinegar. 100% : 20 milliliters

Amount of Italian dressing: 140 milliliters

Step-by-step explanation:

Let's call A the amount of vinegar. 100%

Let's call B the amount of Italian dressing . 12% vinegar

The resulting mixture should have 23% vinegar, and 160 milliliters.

Then we know that the total amount of mixture will be:

$A + B = 160$

Then the total amount of pure antifreeze in the mixture will be:

$A + 0.12B = 0.23 * 160$

$A + 0.12B = 36.8$

Then we have two equations and two unknowns so we solve the system of equations. Multiply the first equation by -1 and add it to the second equation:

$-A -B = -160$

$-A -B = -160$

+

$A + 0.12B = 36.8$

--------------------------------------

$-0.88B = -123.2$

$B = \frac{-123.2}{-0.88}$

$B = 140\ milliliters$

We substitute the value of B into one of the two equations and solve for A.

$A + 140 = 160$

$A = 20\ milliliters$

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3 years ago

In a card game, the probability that you will have a hand with three of a kind is about 6%. The dealer wants to know the probabi

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Answer:

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Step-by-step explanation:

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2 years ago

A pet store sold 49 siamese cats. If the ratio of siamese cats to house cats sold was 7:4, what is the combined amount of cats s

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2 years ago

Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.

marshall27 [118]

Answer:

$\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\ \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right$

Step-by-step explanation:

a) $\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}$

b) $\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}$

c) $\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0$

d) $\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}$

e) $\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2$

f) $\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}$

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2 years ago