![\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 answer is b
Step-by-step explanation:
Answer:
x = 43 degrees; 2x = 86 degrees; x+ 8 = 51
Step-by-step explanation:
x + 2x + x + 8 = 180
4x + 8 = 180
- 8 -8
4x = 172 (divide both sides by 4)
x = 43
Answer with Step-by-step explanation:
We know that the components of velocity are obtained from position as
![u=\frac{dx}{dt}\\\\v=\frac{dy}{dt}](https://tex.z-dn.net/?f=u%3D%5Cfrac%7Bdx%7D%7Bdt%7D%5C%5C%5C%5Cv%3D%5Cfrac%7Bdy%7D%7Bdt%7D)
Using the given values we obtain
![u=\frac{d(2t^2)}{dt}\\\\u=4t](https://tex.z-dn.net/?f=u%3D%5Cfrac%7Bd%282t%5E2%29%7D%7Bdt%7D%5C%5C%5C%5Cu%3D4t)
Similarly
![v=\frac{d(t^2-41)}{dt}\\\\u=2t](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%28t%5E2-41%29%7D%7Bdt%7D%5C%5C%5C%5Cu%3D2t)
The the velocity function can be written as
![\overrightarrow{v}=4t\widehat{i}+2t\widehat{j}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D4t%5Cwidehat%7Bi%7D%2B2t%5Cwidehat%7Bj%7D)
The components of acceleration are obatined from the components of velocity as
![a_{x}=\frac{du}{dt}\\\\a_{y}=\frac{dv}{dt}](https://tex.z-dn.net/?f=a_%7Bx%7D%3D%5Cfrac%7Bdu%7D%7Bdt%7D%5C%5C%5C%5Ca_%7By%7D%3D%5Cfrac%7Bdv%7D%7Bdt%7D)
Using the given values we obtain
![a_x=\frac{d(4t)}{dt}\\\\a_{x}=4](https://tex.z-dn.net/?f=a_x%3D%5Cfrac%7Bd%284t%29%7D%7Bdt%7D%5C%5C%5C%5Ca_%7Bx%7D%3D4)
Similarly
![a_y=\frac{d(2t)}{dt}\\\\a_y=2](https://tex.z-dn.net/?f=a_y%3D%5Cfrac%7Bd%282t%29%7D%7Bdt%7D%5C%5C%5C%5Ca_y%3D2)
The the acceleration function can be written as
![\overrightarrow{a}=4\widehat{i}+2\widehat{j}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Ba%7D%3D4%5Cwidehat%7Bi%7D%2B2%5Cwidehat%7Bj%7D)
Thus at time 't=1' the velocity function becomes
![\overrightarrow{v}=4\widehat{i}+2\widehat{j}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D4%5Cwidehat%7Bi%7D%2B2%5Cwidehat%7Bj%7D)
Thus the component of acceleration in the direction of the given vector
can be found by taking the dot product of the 2 vectors
Thus we get
![v_{r}=\overrightarrow{v}\cdot \overrightarrow{r}\\\\v_{r}=(4\widehat{i}+2\widehat{j})\cdot (\widehat{i}-3\widehat{j})\\\\v_{r}=4-6=-2](https://tex.z-dn.net/?f=v_%7Br%7D%3D%5Coverrightarrow%7Bv%7D%5Ccdot%20%5Coverrightarrow%7Br%7D%5C%5C%5C%5Cv_%7Br%7D%3D%284%5Cwidehat%7Bi%7D%2B2%5Cwidehat%7Bj%7D%29%5Ccdot%20%28%5Cwidehat%7Bi%7D-3%5Cwidehat%7Bj%7D%29%5C%5C%5C%5Cv_%7Br%7D%3D4-6%3D-2)
Similarly the dot product is obtained for acceleration as
![a_{r}=\overrightarrow{a}\cdot \overrightarrow{r}\\\\a_{r}=(4\widehat{i}+2\widehat{j})\cdot (\widehat{i}-3\widehat{j})\\\\a_{r}=4-6=-2](https://tex.z-dn.net/?f=a_%7Br%7D%3D%5Coverrightarrow%7Ba%7D%5Ccdot%20%5Coverrightarrow%7Br%7D%5C%5C%5C%5Ca_%7Br%7D%3D%284%5Cwidehat%7Bi%7D%2B2%5Cwidehat%7Bj%7D%29%5Ccdot%20%28%5Cwidehat%7Bi%7D-3%5Cwidehat%7Bj%7D%29%5C%5C%5C%5Ca_%7Br%7D%3D4-6%3D-2)