1)144 ft
2)6,600 cm
3)1,008 m
4)9,720m
Vertex form is y=a(x-h)^2+k, so we can rearrange to that form...
y=3x^2-6x+2 subtract 2 from both sides
y-2=3x^2-6x divide both sides by 3
(y-2)/3=x^2-2x, halve the linear coefficient, square it, add it to both sides...in this case: (-2/2)^2=1 so
(y-2)/3+1=x^2-2x+1 now the right side is a perfect square
(y-2+3)/3=(x-1)^2
(y+1)/3=(x-1)^2 multiply both sides by 3
y+1=3(x-1)^2 subtract 1 from both sides
y=3(x-1)^2-1 so the vertex is:
(1, -1)
...
Now if you'd like you can commit to memory the vertex point for any parabola so you don't have to do the calculations like what we did above. The vertex of any quadratic (parabola), ax^2+bx+c is:
x= -b/(2a), y= (4ac-b^2)/(4a)
Then you will always be able to do a quick calculation of the vertex :)
Area= base x altura dividido entre 2
Answer:
No, the smallest ratio between a hypotenuse and a leg is in the 45/45/90 special case. In this case, the ratio between the hypotenuse and either legs is
√2:1
If the leg is 10 inches long, the smallest the hypotenuse can be is 10√2.
The equation is y=1x-2
Step-by-step explanation:
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