RainbowSquirrel53

? Where's the array?

I'm guessing the question is asking for the population ten years ago.

Since the population has decreased by 30%, the population now (350,000) is 70% = 7/10 of the original population. So then,

$(\text{population 10 years ago})\cdot\frac{7}{10}=350,000$.

Multiplying both sides by 10/7 to cancel out the fraction gives

$\text{population 10 years ago} = 350,000\cdot\frac{10}{7}=\boxed{500,000}$

$x^2=-64,x=±\sqrt{-64},x=\boxed{-8i,8i}$.

You can't find the square root of that... Do you mean sqrt(-1) and sqrt(4)? sqrt(-1) = i and sqrt(4) = 2

-8 < 3x + 1 < 7

-9 < 3x < 6

-3 < x < 2

that's why I'm asking this question! How do you do it?

I don't understand what you're doing there. Once you get 57, you basically start over the process. You do the same thing you did to 12345 to 57.

$\frac{888}{1000}=\boxed{\frac{111}{125}}$

I think it's asking you to do the same 57, which would turn into 5*2+7=17, and then do it on 17, which would turn into 1*2+7=9.

This is an arithmetic sequence with common difference 7, so the next two numbers are 66 and 73.

$x=\boxed{-\frac{2}{3},-4}$

Seriously? Test question? Welp your test is over anyways, so the answer is $\boxed{7}$ hours.

The food cost $\boxed{\$90}$, and the drinks cost $\boxed{\$18}$. Each girl owes $\boxed{\$36}$.

How many values of $x$ satisfy both of the following conditions?

$(a)$ $x$ is a multiple of $7,$ not necessarily positive.

$(b)$ $x^2$ is less than $500$.

If $x^2$ is $500,$ that means $-22\le x\le22$. Also, since $x$ is a multiple of $7,$ we can list all the multiples of 7 between -22 and 22 out.

-21, -14, -7, 0, 7, 14, 21

Therefore, since there are 7 terms in the list, the answer is $\boxed{7}$.

Press the 2nd key and then the test/math key.

I don't get why this is "college" algebra, but the answer is 8.5.