Answer:
Ben needs to get a 99% on his second test
Step-by-step explanation:
The average is the sum of both test scores, divided by the total number of test scores (2).
Let x by the grade Ben must receive on his second test.
(89 + x)/2 = 94
Multiply by 2 on both sides
89 + x = 188
Subtract 89 on both sides
x = 99
Check work:
(89 + 99)/2 = 188/2 = 99
Since the limit becomes the undetermined form
![\displaystyle \lim_{x\to 1} \dfrac{x^3-2x^2+3x-2}{2x^4-3x+1} \to \dfrac{0}{0}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%201%7D%20%5Cdfrac%7Bx%5E3-2x%5E2%2B3x-2%7D%7B2x%5E4-3x%2B1%7D%20%5Cto%20%5Cdfrac%7B0%7D%7B0%7D%20)
it means that both polynomials have a root at
. So, we can fact both numerator and denominator:
![x^3-2x^2+3x-2 = (x-1)(x^2-x+2)](https://tex.z-dn.net/?f=%20x%5E3-2x%5E2%2B3x-2%20%3D%20%28x-1%29%28x%5E2-x%2B2%29%20)
![2x^4-3x+1 = (x-1)(2x^3+2x^2+2x-1)](https://tex.z-dn.net/?f=%202x%5E4-3x%2B1%20%3D%20%28x-1%29%282x%5E3%2B2x%5E2%2B2x-1%29%20)
So, the fraction becomes
![\dfrac{(x-1)(x^2-x+2)}{(x-1)(2x^3+2x^2+2x-1)} = \dfrac{x^2-x+2}{2x^3+2x^2+2x-1}](https://tex.z-dn.net/?f=%20%5Cdfrac%7B%28x-1%29%28x%5E2-x%2B2%29%7D%7B%28x-1%29%282x%5E3%2B2x%5E2%2B2x-1%29%7D%20%3D%20%5Cdfrac%7Bx%5E2-x%2B2%7D%7B2x%5E3%2B2x%5E2%2B2x-1%7D%20)
Now, as x approaches 1, you have no problems anymore:
![\displaystyle \lim_{x\to 1} \dfrac{x^2-x+2}{2x^3+2x^2+2x-1} \to \dfrac{2}{5}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%201%7D%20%5Cdfrac%7Bx%5E2-x%2B2%7D%7B2x%5E3%2B2x%5E2%2B2x-1%7D%20%5Cto%20%5Cdfrac%7B2%7D%7B5%7D%20)
First you have to distribute. -3y +12=18. -3y=6. y= -2.
The Answer is False. It is false because on line has a slope of 0 and the other has an undefined slope.
Hope this helps, Good luck!
Answer:
ff
Step-by-step explanation: