![\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.
Step-by-step explanation:
let necklace : x
let earrings : y
x+y=144 -1st eqn
6x+5y=825 -2nd eqn
y=144-X -3rd eqn
substitute eqn into 2nd eqn
6x+5(144-x)= 825
6x +720-5X =825 (expanded)
X = 825-720
X =105
substitute "X=105" into 1st eqn
(105)+y=144
y= 144-105
y=39
1 necklace = 105$
1 pair of earrings = 39$
(to check back: substitute the value into the original eqn)
Answer:
<em><u>1 and 5</u></em>
Step-by-step explanation:
The squares have a side length of 10 and 1 square side is the radius of the half-circles. Since there are two half-circles, find the circumference for one full circle:

Insert the radius:

Simplify pi:

Simplify multiplication:

The circumference of the circles is 62.8. Now find the perimeter of the exposed squares with side length 10. There are 4 exposed sides, which equals one square. Find the perimeter:

Add the perimeter of the circle and the square together:

Now see which of the options gives you the perimeter:
1.
****
2. 
3. 
4. 
5.
****
Finito.
When you simplify the expression 1/1+cot^2x the final product is C. sin^2 (x).
Hope this helps!
Answer:
0
Step-by-step explanation:
the highest the stone will ever be is in your hand after you throw it the rock will automatically start falling more and more down towards the river which is below you.
Hopefully this helps!