![\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.
Answer: the solution becomes 3=y+1-1
Step-by-step explanation:
The first step is to combine like terms meaning you subtract 1-1. Then you are left with 3=y.
Hi!
A rectangle can sometimes be a rhombus if it has sides of all equal lengths. If a rectangle was a rhombus, it would also be a square. Furthermore, a square is always a rhombus.
Hope this helps! :)
Answer:
3x^2+2x+15
Step-by-step explanation:
3x^2+2x+2+(4x2)+5
use pemdas
The increasing order of the horizontal widths of their asymptote rectangles is dependent on the values gotten from y = ± x.
<h3>What is a Hyperbola?</h3>
This is defined as a two-branched open curve formed by the intersection of a plane perpendicular to the bases of a double cone.
The rectangular hyperbola has two asymptotes which are defined as y = ± x in this scenario.
Read more about Hyperbola here brainly.com/question/3351710
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