Answer:
y = (-7/3)(x - 9)
Step-by-step explanation:
<em><u>(12, - 7)</u></em>
y = a(x - h)^2
-7 = a(12 - h)^2
- 7 = a(144 - 24x + h^2)
<em><u>(9,0)</u></em>
0 = a(x - h)^2
0 = a(9 - h)^2
0 = a(81 - 18h + h^2)
From (9,0) we can conclude that
a = 0
or
(9 - h)^2 = 0
Let's try the second possibility.
Take the square root of both sides.
9 - h = sqrt(0)
Add h to both sides
9 - h = 0
h = 9
So now what we have is
y = a(x - 9)
Use the first equation to get a
-7 = a(12 - 9)
-7 = a(3)
-7/3 = a
Answer
y = (-7/3)(x - 9)
Answer:
they would have saved 35 dollars
Step-by-step explanation:
Ralph on his own would have paid 105 and jeff 140 but together with the discount they could save $35 by paying 210 in total with the discount.
Answer:
Step-by-step explanation:
Given two upward facing parabolas with equations
The two intersect at
=
x=
area enclosed by them is given by
A=![\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx](https://tex.z-dn.net/?f=%5Cint_%7B-%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%7D%5E%7B%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%7D%5Cleft%20%5B%20%5Cleft%20%28%20x%5E2%2B2%5Cright%20%29-%5Cleft%20%28%206x%5E2%5Cright%20%29%20%5Cright%20%5Ddx)
A=
A=![4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ]](https://tex.z-dn.net/?f=4%5Cleft%20%5B%20%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%20%5Cright%20%5D-%5Cfrac%7B5%7D%7B3%7D%5Cleft%20%5B%20%5Cleft%20%28%20%5Cfrac%7B2%7D%7B5%7D%5Cright%20%29%5E%5Cfrac%7B3%7D%7B2%7D-%5Cleft%20%28%20-%5Cfrac%7B2%7D%7B5%7D%5Cright%20%29%5E%5Cfrac%7B3%7D%7B2%7D%20%5Cright%20%5D)
A=
You would have to make a point at 6/3 and go down one and to the left 7... then your answer would be -1/2 sooo D=-1 hope this helps
Answer:
<D = 82°
<E = 49°
<F = 49°
Step-by-step explanation:
This is an isosceles triangle because two of the sides are the same length which means the 2 bottom angles are the exact same which means we can set them equal to each other to solve for it.
