1/8 of the cake has not been eaten.
Answer:
No, no puedo ayudarte aunque te deseo buena suerte
Answer:
22 cents
Step-by-step explanation:
Dividing 75 by 3 gets you 25. "Less than" as in subtracting, so subtract 3 from 25 and you get 22.
Answer:
We take v in the vector space V different from 0. Since v is not 0, then
![< v,v>_1 \, >0 \, ;, _2 \, >0](https://tex.z-dn.net/?f=%3C%20v%2Cv%3E_1%20%5C%2C%20%3E0%20%5C%2C%20%3B%2C%20%3Cv%2Cv%3E_2%20%5C%2C%20%3E0)
Lets take k>0 such that
![_2 \, = k*\, _1](https://tex.z-dn.net/?f=%20%3Cv%2Cv%3E_2%20%5C%2C%20%3D%20k%2A%5C%2C%20%3Cv%2Cv%3E_1%20)
Now, we take any vector w. We want to show that
.
Since v is any non-zero vector, then this will prove that
for any vectors a,b. The reason is that for any vector different from 0, lets name it x, there will exist a constant
such that
for any y (this is for the same reason a constant exists for v). Since y can be anything, then it can be v. But that means that
, because v also has its constant k.
Now, lets show that
. Lets take a constant c such that
. We have that
![0 = _2 - c* _2 = _2](https://tex.z-dn.net/?f=%200%20%3D%20%3Cv%2Cw%3E_2%20-%20c%2A%20%3Cv%2Cv%3E_2%20%3D%20%3Cv%2C%20w-cv%3E_2%20)
Thus
![_1 = 0](https://tex.z-dn.net/?f=%20%3Cv%2Cw-cv%3E_1%20%3D%200%20)
Which means that
![0 = _1 = _1 - c_1](https://tex.z-dn.net/?f=%200%20%3D%20%3Cv%2Cw-cv%3E_1%20%3D%20%3Cv%2Cw%3E_1%20-%20c%3Cv%2Cv%3E_1%20)
Which means that
. As a consequence
![_2 = c _2 = ck*_1 = k * _1](https://tex.z-dn.net/?f=%20%3Cv%2Cw%3E_2%20%3D%20c%20%3Cv%2Cv%3E_2%20%3D%20ck%2A%3Cv%2Cv%3E_1%20%3D%20k%20%2A%20%3Cv%2Cw%3E_1%20)
Which proves what we were looking for.
The time will it take the runners to cover 1/A mile
t= 5904.6 sec
<h3>What time will it take the runners to cover 1/A mile?</h3>
Joel takes 63 sec for 14-mile long track
speed of joel
![\frac{14}{1.05} \mathrm{mi} / \mathrm{min}=13.33 \mathrm{mi} / \mathrm{min}](https://tex.z-dn.net/?f=%5Cfrac%7B14%7D%7B1.05%7D%20%5Cmathrm%7Bmi%7D%20%2F%20%5Cmathrm%7Bmin%7D%3D13.33%20%5Cmathrm%7Bmi%7D%20%2F%20%5Cmathrm%7Bmin%7D)
Jason takes 57 \mathrm{sec} for 14 mi track
so Jason's speed is
![\frac{14}{0.95} \mathrm{mi} / \mathrm{min}=14.74 \mathrm{mi} / \mathrm{min}](https://tex.z-dn.net/?f=%5Cfrac%7B14%7D%7B0.95%7D%20%5Cmathrm%7Bmi%7D%20%2F%20%5Cmathrm%7Bmin%7D%3D14.74%20%5Cmathrm%7Bmi%7D%20%2F%20%5Cmathrm%7Bmin%7D)
therefore
if both start from the same spot in opposite direction then
![\frac{x}{13.33}=\frac{14-x}{14.74}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B13.33%7D%3D%5Cfrac%7B14-x%7D%7B14.74%7D)
where x is the distance covered by Joel
then![x=131.91$ miles](https://tex.z-dn.net/?f=x%3D131.91%24%20miles)
time required
![t=\frac{131.91}{13.33}=98.41 \mathrm{~min} 5904.6 sec](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B131.91%7D%7B13.33%7D%3D98.41%20%5Cmathrm%7B~min%7D%205904.6%20sec)
Read more about time
brainly.com/question/28050940
#SPJ1