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9.5:12.6
That is the answer, btw the 6 is repeating
Answer:
The answer is C.
There is a horizontal line test but the question asks to use the vertical line test. Using the horizontal line test you would get that the function is not one to one but using the vertical line test you would get its one to one.
Step-by-step explanation:
![2 {x}^{2} + 7 + 10 - ( {x}^{2} + 2x - 9) \\ 2 {x}^{2} + 7 + 10 - {x }^{2} - 2x + 9 \\ {x}^{2} - 2x + 26](https://tex.z-dn.net/?f=2%20%7Bx%7D%5E%7B2%7D%20%20%2B%207%20%2B%2010%20-%20%28%20%7Bx%7D%5E%7B2%7D%20%2B%202x%20-%209%29%20%5C%5C%202%20%7Bx%7D%5E%7B2%7D%20%20%20%2B%207%20%2B%2010%20-%20%20%7Bx%20%7D%5E%7B2%7D%20%20-%202x%20%20%2B%209%20%5C%5C%20%20%7Bx%7D%5E%7B2%7D%20%20-%202x%20%2B%2026)
Answer:
Given,
ABC is triangle with vertices A(-2,1), B(2,2) and C(6, -2)
Slope of BC(m_1) =![\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{-2-2}](https://tex.z-dn.net/?f=%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%3D%5Cfrac%7B6-2%7D%7B-2-2%7D)
Since AD is perpendicular to BC
let slope of BC be (
).
We have for perpendicular
![m_1*m_2=-1](https://tex.z-dn.net/?f=m_1%2Am_2%3D-1)
![-1*m_2=-1](https://tex.z-dn.net/?f=-1%2Am_2%3D-1)
![m_2=1](https://tex.z-dn.net/?f=m_2%3D1)
Now equation of line passing through point A(-2,1) is
![y-y_1=m_2(x-x_1)](https://tex.z-dn.net/?f=y-y_1%3Dm_2%28x-x_1%29)
y-1=1(x-(-2))
y-1=x+2
y=x+2+1
y=x+3 which is a required equation of the altitude of AD drawn from A to BC.
Answer:
Trevor is just under halfway through his homework
Step-by-step explanation:
The numbers of homework are 12
Trevor did 5 of them. Halfway is 6