Answer:
C-$6.00 monthly fee and $1.25 per video game
Step-by-step explanation:
Answer:
Step-by-step explanation:
Complement of the event
![= 1 - \frac{1}{9} \\ = \frac{9 - 1}{9} \\ = \frac{8}{9} \\](https://tex.z-dn.net/?f=%20%3D%201%20-%20%20%5Cfrac%7B1%7D%7B9%7D%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B9%20-%201%7D%7B9%7D%20%20%5C%5C%20%20%3D%20%20%5Cfrac%7B8%7D%7B9%7D%20%20%5C%5C%20)
Answer:
w<-6
Step-by-step explanation:
11w+99<33
-99
11w<-66/11
✩ Answer:
✧・゚: *✧・゚:*✧・゚: *✧・゚:*✧・゚: *✧・゚:*
✩ Step-by-step explanation:
✧・゚: *✧・゚:*✧・゚: *✧・゚:*✧・゚: *✧・゚:*
✺ Divide both sides by −3. Since −3 is negative, the inequality direction is changed:
- -3 ÷ -3 = Cancels Out
- Turns into:
![\frac{-9}{-3} >5x-2](https://tex.z-dn.net/?f=%5Cfrac%7B-9%7D%7B-3%7D%20%3E5x-2)
✺ Divide −9 by −3 to get 3. Since they are both negatives it turns into a positive:
- -9 ÷ -3 = 3
- Turns into:
![3 > 5x - 2](https://tex.z-dn.net/?f=3%20%3E%205x%20-%202)
✺ Swap sides so that all variable terms are on the left hand side. This changes the sign direction once more:
- Turns into:
![5x-2](https://tex.z-dn.net/?f=5x-2%3C3)
✺ Add 2 to both sides:
- Turns into:
![5x](https://tex.z-dn.net/?f=5x%3C3%2B2)
✺ Add 3 and 2 to get 5:
- 3 + 2 = 5
- Turns into:
![5x](https://tex.z-dn.net/?f=5x%3C5)
✺ Divide both sides by 5. Since 5 is positive, the inequality direction remains the same:
- 5 ÷ 5 = Cancels Out
- Turns into:
![x](https://tex.z-dn.net/?f=x%3C%5Cfrac%7B5%7D%7B5%7D)
✺ Divide 5 by 5 to get 1:
✺ So your answer is B,
.
(Number line is below.)
With
![\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20r%28t%29%3D4t%5C%2C%5Cvec%5Cimath%2B6t%5C%2C%5Cvec%5Cjmath-t%5E2%5C%2C%5Cvec%20k)
we have
![\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cvec%20r%3D%284%5C%2C%5Cvec%5Cimath%2B6%5C%2C%5Cvec%5Cjmath-2t%5C%2C%5Cvec%20k%29%5C%2C%5Cmathrm%20dt)
The vector field evaluated over this parameterization is
![\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20f%28x%2Cy%2Cz%29%3D%5Cvec%20f%28x%28t%29%2Cy%28t%29%2Cz%28t%29%29%3D4t%5C%2C%5Cvec%5Cimath%2Bt%5E2%5C%2C%5Cvec%5Cjmath%2B6t%5C%2C%5Cvec%20k)
so the line integral is
![\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B-1%7D%5E1%284t%5C%2C%5Cvec%5Cimath%2Bt%5E2%5C%2C%5Cvec%5Cjmath%2B6t%5C%2C%5Cvec%20k%29%5Ccdot%284%5C%2C%5Cvec%5Cimath%2B6%5C%2C%5Cvec%5Cjmath-2t%5C%2C%5Cvec%20k%29%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_%7B-1%7D%5E1%2816t%2B6t%5E2-12t%5E2%29%5C%2C%5Cmathrm%20dt%3D-4)