1 year ago

Y=x^2 is y a function of x

1 answer:

4 0

Yes y=x^2 is a function because Using the definition of a function, y=x^2 is a function because for every x-value in your domain, you will only get 1 unique y-value. A quick check you can do is to use the vertical line test

You might be interested in

What are the next 3 terms in this:O,T,T,F,F

Rasek [7]

Answer:

Step-by-step explanation:

S, S, E.

Come on, do I have to spell it out for you? :)

8 0

2 years ago

2. Which of the following is the correct interpretation of confidence interval? (circle one)

Kobotan [32]

B. The answer is b hehehehhehsheus

6 0

2 years ago

Find the missing point of the following parallelogram.

zepelin [54]

Answer:

(4,5)

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Solve h = −16t2 + 36t + 4.

kvasek [131]

I can do each of the things I asked above.
So, let's change this into vertex form:
$h=-16t^2+36t+4$
$h=(-16t^2+36t)+4$
$h=-16(t^2-2.25t)+4$
$h=-16(t^2-2.25t+1.27-1.27)+4$
$h=-16(t^2-2.25t+1.27)+20.32+4$
$h=-16(t-1.125)^2+24.32$
The vertex is at (1.125,24.32)
<span>Answers may vary due to rounding

</span>Factored Form:
$h = -16t^2 + 36t + 4$
$h = -4\left(4t^2-9t-1\right)$

Quadratic Formula:
$x = \frac{-b +/- \sqrt{b^2-4(a)(c)}}{2a}$
<span>h = -16t^2 + 36t + 4
</span>a = -16 b = 36 c = 4
$h = \frac{-(36) +/- \sqrt{(36)^2-4(-16)(4)}}{2(-16)}$
$h = \frac{-36 +/- \sqrt{1552}}{-32}$
$h = ≈-0.11$
$h = ≈2.36$

5 0

2 years ago

Help math pls its a drag and drop

STALIN [3.7K]

A goes in stock price decreased
B goes in stock price increased
C goes in stock price remained constant
D goes in stock price decreased
E goes is stock price increased

Hope this helps!

7 0

2 years ago

Read 2 more answers

Y=x^2 is y a function of x​

Y=x^2 is y a function of x