Please help me with these math questions. NO LINKS PLEASE!! Will give brainliest if correct!! (Pictures attached) :)

Mathematics

1 answer:

alexgriva [62]3 years ago

8 0

Answer:

See below

Step-by-step explanation:

Domain: negative infinity, positive infinity

Range: -392, positive infinity

Y-intercept: (0, -150)

X-intercepts: (25, 0) (-3, 0)

Axis of symmetry: x = 11

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The dot plots below show the weights of the players of two teams:

Nadusha1986 [10]

Answer:B

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Somebody please help so I can pass, please

ASHA 777 [7]

First, we are going to find the vertex of our quadratic. Remember that to find the vertex $(h,k)$ of a quadratic equation of the form $y=a x^{2} +bx+c$ , we use the vertex formula $h= \frac{-b}{2a}$ , and then, we evaluate our equation at $h$ to find $k$ .

We now from our quadratic that $a=2$ and $b=-32$ , so lets use our formula:
$h= \frac{-b}{2a}$
$h= \frac{-(-32)}{2(2)}$
$h= \frac{32}{4}$
$h=8$
Now we can evaluate our quadratic at 8 to find $k$ :
$k=2(8)^2-32(8)+56$
$k=2(64)-256+56$
$k=128-200$
$k=-72$
So the vertex of our function is (8,-72)

Next, we are going to use the vertex to rewrite our quadratic equation:
$y=a(x-h)^2+k$
$y=2(x-8)^2+(-72)$
$y=2(x-8)^2-72$
The x-coordinate of the minimum will be the x-coordinate of the vertex; in other words: 8.

We can conclude that:
The rewritten equation is $y=2(x-8)^2-72$
The x-coordinate of the minimum is 8