![\bf 400,000,000\implies 4\times 10^8 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\textit{desktop users}}{\textit{mobile users}}\qquad \qquad \cfrac{1.2\times 10^9}{4\times 10^8}\implies \cfrac{12\times 10^8}{4\times 10^8}\implies \cfrac{12}{4}\times\cfrac{10^8}{10^8}\implies \cfrac{3}{1}](https://tex.z-dn.net/?f=%5Cbf%20400%2C000%2C000%5Cimplies%204%5Ctimes%2010%5E8%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B%5Ctextit%7Bdesktop%20users%7D%7D%7B%5Ctextit%7Bmobile%20users%7D%7D%5Cqquad%20%5Cqquad%20%5Ccfrac%7B1.2%5Ctimes%2010%5E9%7D%7B4%5Ctimes%2010%5E8%7D%5Cimplies%20%5Ccfrac%7B12%5Ctimes%2010%5E8%7D%7B4%5Ctimes%2010%5E8%7D%5Cimplies%20%5Ccfrac%7B12%7D%7B4%7D%5Ctimes%5Ccfrac%7B10%5E8%7D%7B10%5E8%7D%5Cimplies%20%5Ccfrac%7B3%7D%7B1%7D)
3 : 1, or 3 to 1, thus 3 times as many.
The domain is all the x values (range is y values). Remember to put the x values in numerical order.
So: Domain = {-5, -5, -2, 3}.
he solution set is
{
x
∣
x
>
1
}
.
Explanation
For each of these inequalities, there will be a set of
x
-values that make them true. For example, it's pretty clear that large values of
x
(like 1,000) work for both, and negative values (like -1,000) will not work for either.
Since we're asked to solve a "this OR that" pair of inequalities, what we'd like to know are all the
x
-values that will work for at least one of them. To do this, we solve both inequalities for
x
, and then overlap the two solution set
For dividing fractions you should always use the KCF method: Keep Change Flip
6/10 divided by 12/7
Kcf method:
6/10*7/12= 42/120
42/120 in simplest form will be 7/20
So the answer is A
Answer:
idk
Step-by-step explanation:
