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3 years ago

Does anyone know these answers? 50 pts to you if u do.....don't steal from me.

Mathematics

2 answers:

MArishka [77]3 years ago

6 0

Answer:

y = -2x + 3

y = -x + 10

y = 4x + 10

y = 3/2x - 2

y = -1/2x - 1

y = x + 5

y = 25x + 5

y = .99x + 1.99

y = x - 2

y = -1.50 + 1.50

y = 42x + 27

y = -1/3x - 2

*answering for 2nd time because my first got deleted. eek.

snow_lady [41]3 years ago

3 0

y = -2x + 3

y = -x + 10

y = 4x + 10

y = 3/2x - 2

y = -1/2x - 1

y = x + 5

y = 25x + 5

y = .99x + 1.99

y = x - 2

y = -1.50 + 1.50

y = 42x + 27

y = -1/3x - 2

now i am in my phone can see the photo

sometimes in the computer don't appear the picture

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1 year ago

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Answer:

$x=1+\sqrt{3}$ and $x=1-\sqrt{3}$

Step-by-step explanation:

We have been given the parabola with vertex (1, -9) and y intercept at (0, -6).

Now we need to find the x-intercepts of that parabola. So first we begin by finding the equation of parabola using vertex formula:

$y=a\left(x-h\right)^2+k$

Vertex for this formula is given by (h,k)

Compare that with given vertex (1,-9), we get: h=1, k=-9

So plug these into vertex formula:

$y=a\left(x-1\right)^2-9$ ...(i)

Plug given point (0, -6). into (i)

$-6=a\left(0-1\right)^2-9$

$-6=a\left(-1\right)^2-9$

$-6=a\left(1\right)-9$

$-6+9=a$

$3=a$

Plug a=3 into (i)

$y=3\left(x-1\right)^2-9$

Now to find x-intercept, we just plug y=0 and solve for x

$y=3\left(x-1\right)^2-9$

$0=3\left(x-1\right)^2-9$

$9=3\left(x-1\right)^2$

$3=\left(x-1\right)^2$

$\pm \sqrt{3}=x-1$

$1+\pm \sqrt{3}=x$

Hence final answer are

$x=1+\sqrt{3}$ and $x=1-\sqrt{3}$

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3 years ago