The axis of symmetry of a quadratic function
![f(x)=ax^2+bx+c](https://tex.z-dn.net/?f=f%28x%29%3Dax%5E2%2Bbx%2Bc)
is given by the equation
![x=h](https://tex.z-dn.net/?f=x%3Dh)
, where h is the x-coordinate of the vertex and is equal to
![\frac{-b}{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b%7D%7B2a%7D)
.
1.
![f(x)=4x^2-1 \\ a=4 \\ b=0 \\ \Downarrow \\ h=\frac{-0}{2 \times 4}=0 \\ \\ \hbox{the axis of symmetry:} \\ x=0](https://tex.z-dn.net/?f=f%28x%29%3D4x%5E2-1%20%5C%5C%0Aa%3D4%20%5C%5C%20b%3D0%20%5C%5C%20%5CDownarrow%20%5C%5C%0Ah%3D%5Cfrac%7B-0%7D%7B2%20%5Ctimes%204%7D%3D0%20%5C%5C%20%5C%5C%0A%5Chbox%7Bthe%20axis%20of%20symmetry%3A%7D%20%5C%5C%0Ax%3D0)
2.
![g(x)=x^2-8x+5 \\ a=1 \\ b=-8 \\ \Downarrow \\ h=\frac{-(-8)}{2 \times 1}=\frac{8}{2}=4 \\ \\ \hbox{the axis of symmetry:} \\ x=4](https://tex.z-dn.net/?f=g%28x%29%3Dx%5E2-8x%2B5%20%5C%5C%0Aa%3D1%20%5C%5C%20b%3D-8%20%5C%5C%20%5CDownarrow%20%5C%5C%20h%3D%5Cfrac%7B-%28-8%29%7D%7B2%20%5Ctimes%201%7D%3D%5Cfrac%7B8%7D%7B2%7D%3D4%20%5C%5C%20%5C%5C%0A%5Chbox%7Bthe%20axis%20of%20symmetry%3A%7D%20%5C%5C%0Ax%3D4)
3.
![h(x)=-3x^2-12x+1 \\ a=-3 \\ b=-12 \\ \Downarrow \\ h=\frac{-(-12)}{2 \times (-3)}=\frac{12}{-6}=-2 \\ \\ \hbox{the axis of symmetry: \\ x=-2](https://tex.z-dn.net/?f=h%28x%29%3D-3x%5E2-12x%2B1%20%5C%5C%0Aa%3D-3%20%5C%5C%20b%3D-12%20%5C%5C%20%5CDownarrow%20%5C%5C%20h%3D%5Cfrac%7B-%28-12%29%7D%7B2%20%5Ctimes%20%28-3%29%7D%3D%5Cfrac%7B12%7D%7B-6%7D%3D-2%20%5C%5C%20%5C%5C%0A%5Chbox%7Bthe%20axis%20of%20symmetry%3A%20%5C%5C%0Ax%3D-2)
The functions ranked from least to greatest based on their axis of symmetry: h(x), f(x), g(x).
Answer:
8
Step-by-step explanation:
hypotenuse^2= leg^2 + leg^2
10^2= 6^2 + x^2
100= 36 + x^2
100-36= x^2
64= x^2
√64=√x^2
8= x
The answer to your question is 2/3 because there are 2 cups and 2/3 of buttermilk meaning 33.3333.. so it would be 20/30 then 2/3
Answer:
91.8 ft
Step-by-step explanation:
So we can talk about the diagram, let's name a couple of points. The base of the tree is point T, and the top of the tree is point H. We want to find the length of TH given the length AB and the angles HAT and ABT.
The tangent function is useful here. By its definition, we know that ...
TA/BA = tan(∠ABT)
and
TH/TA = tan(∠HAT)
Then we can solve for TH by substituting for TA. From the first equation, ...
TA = BA·tan(∠ABT)
From the second equation, ...
TH = TA·tan(∠HAT) = (BA·tan(∠ABT))·tan(∠HAT)
Filling in the values, we get ...
TH = (24.8 ft)tan(87.3°)tan(9.9°) ≈ 91.8 ft
The height <em>h</em> of the tree is about 91.8 ft.