I think it’s -1/2 because you would have to do the equation the opposite way so whenever it says divide you need to multiply etc.
Consider this option:
1. rule:
![V_{con}= \frac{1}{3} \pi r^2h, \ where \ r-radius \ of \ the \ base,h-height \ of \ the \ cone](https://tex.z-dn.net/?f=V_%7Bcon%7D%3D%20%5Cfrac%7B1%7D%7B3%7D%20%20%5Cpi%20r%5E2h%2C%20%5C%20where%20%5C%20r-radius%20%5C%20of%20%5C%20the%20%5C%20base%2Ch-height%20%5C%20of%20%5C%20the%20%5C%20cone)
2. If r=1 ft., h=2ft., then
V=1/3 * 1*2*π=2/3 π (units)³.
Answer: A. ![h(x)=16x^2-50x+38](https://tex.z-dn.net/?f=h%28x%29%3D16x%5E2-50x%2B38)
Step-by-step explanation:
Given: Francesca graphs the function
, George graphs the function
.
Harold graphs the function
.
i.e.
[put value of f(x)]
![=4(2x-3)^2-(2x-3)-1\\\\= 4((2x)^2-2(2x)(3)+3^2)-2x+3-1\ \ [(a-b)^2=a^2-2ab+b^2]\\\\= 4(4x^2-12x+9)-2x+2\\\\=16x^2-48x+36-2x+2\\\\=16x^2-50x+38](https://tex.z-dn.net/?f=%3D4%282x-3%29%5E2-%282x-3%29-1%5C%5C%5C%5C%3D%204%28%282x%29%5E2-2%282x%29%283%29%2B3%5E2%29-2x%2B3-1%5C%20%5C%20%5B%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2%5D%5C%5C%5C%5C%3D%204%284x%5E2-12x%2B9%29-2x%2B2%5C%5C%5C%5C%3D16x%5E2-48x%2B36-2x%2B2%5C%5C%5C%5C%3D16x%5E2-50x%2B38)
So, Harold graph ![h(x)=16x^2-50x+38](https://tex.z-dn.net/?f=h%28x%29%3D16x%5E2-50x%2B38)
Hence, the correct option is A. ![h(x)=16x^2-50x+38](https://tex.z-dn.net/?f=h%28x%29%3D16x%5E2-50x%2B38)
Answer:
2.85 and 3.45
Step-by-step explanation:
it gpea by 20s I think