Tabitha wants to hang a painting in a gallery. The painting and frame must have an area of 58 square feet. The painting is 7 feet wide by 8 feet long. Which quadratic equation can be used to determine the thickness of the frame, x?
Image of a frame 7 feet wide by 8 feet long, with an of x on the bottom and right sides
x2 + 15x − 2 = 0
x2 + 15x + 58 = 0
4x2 + 30x − 2 = 0
4x2 + 30x + 58 = 0

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- katieb

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A graph of a quadratic function is shown below.
graph of y equals negative 2 times x squared plus 40 x plus 600
What is the vertex of the parabola shown?
(30, 0)
(800, 10)
(0, 600)
(10, 800)

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## More answers

- help_people

- help_people

- jim_thompson5910

Please post questions one at a time. Which one is giving you more trouble?

- help_people

umm the seocnd one and srry :( @jim_thompson5910

- jim_thompson5910

that's ok
do you see how the given equation matches up with the form ax^2 + bx + c ?

- help_people

yes

- help_people

wiat not not for the second one

- jim_thompson5910

|dw:1442708862589:dw|

- jim_thompson5910

|dw:1442708890735:dw|

- jim_thompson5910

|dw:1442708908746:dw|

- jim_thompson5910

|dw:1442708921034:dw|

- jim_thompson5910

|dw:1442708932348:dw|

- help_people

ok yes

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- jim_thompson5910

so if a = -2 and b = 40, we can plug them into \[\Large h = -\frac{b}{2a}\]
what is the value of h when you do this?

- help_people

h=-40/2(-2)
h=10

- jim_thompson5910

very good

- jim_thompson5910

what we do next is plug that value (10) in for x
\[\Large y =-2x^2 + 40x + 600\]
\[\Large y =-2(10)^2 + 40(10) + 600\]
\[\Large y = ???\]

- help_people

800 @jim_thompson5910

- help_people

800

- jim_thompson5910

good

- jim_thompson5910

h = 10 and k = 800 making the vertex to be (10,800)

- jim_thompson5910

in general, the vertex is (h,k)

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- jim_thompson5910

I'm already here and responding. There's no need to keep tagging me like that.

- help_people

so the answer is (!0,800)? could you help w/ the other and i will repost

- jim_thompson5910

yeah `(10,800)`

- help_people

ty could you help w/ 1 more?

- help_people

ty could you help w/ 1more?

- jim_thompson5910

sure, post as a new question and tag me in it

- help_people

ok !

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