Pls help

1 answer:

6 0

Answer: both

Unless you mistyped the first number and it’s actually -3 instead of 3 than b is the answer.

To see if they make a function you can graph all the coordinates and do the line test. Or you can look at all the numbers and see if any of the ordered pair have repeating x values. If they do, it’s not a function.

Remember, it’s (x,y) and if any X values repeat, it’s NOT a function

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Angelina_Jolie [31]

Answer:

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8 0

3 years ago

How do you find experimental probability?

OleMash [197]

Experimental probability is doing a test and basing probability on that. Like if you roll a dice 100 times and 6 comes up 21 times, the experimental probability is 21%.

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4 years ago

19. A scale drawing of a doctor's office is shown. What are the

Digiron [165]

Answer: 40 by 60

Step-by-step explanation:

You are given a scale of 1 inches=20 ft

For the 2 inch dimension, you get:

2*20=40

For the 3 inch dimension, you get:

3*20=60

Therefore, the real dimensions of the doctors office from 2 in. by 3 in. are respectively 40 ft, by 60 ft.

Hope this helped!

6 0

4 years ago

Read 2 more answers

Ryan has a bank account balance of −$17.62. As soon as he realizes this, he deposits $26.50 in the account.

nirvana33 [79]

Answer:

$466.93

Step-by-step explanation:

I multiplied by 26.50 & 17.62 and that got me $466.93. Or you can divide that into 17.62 = $1.50

4 0

3 years ago

The position function of a particle in rectilinear motion is given by s(t) = 2t^3 + 21t^2 + 60t + 3 for t ≥ 0 with t measured in

natta225 [31]

As the fellow above said, the particle does make a U-turn at the vertex, and we can find that out by getting the derivative of s(t) and zeroing it out to get the point of the horizontal tangent.

Bearing in mind that ds/dt is really v(t) or the velocity equation, so when we zero out the ds/dt, is the same as saying the velocity went to 0, since it got zeroed out, and then the particle goes over the vertex and changes direction.

$\bf s(t)=2t^3-21t^2+60t+3\implies \boxed{\cfrac{ds}{dt}=6t^2-42t+60}\leftarrow v(t)
\\\\\\
0=6t^2-42t+60\implies 0=t^2-7t+10\implies 0=(t-5)(t-2)
\\\\\\
t=
\begin{cases}
5\\
2
\end{cases}\impliedby \textit{it makes the first turn at the \underline{2 second}}
\\\\\\
\textit{now let's find the acceleration equation, a(t)}
\\\\\\
\boxed{\cfrac{d^2s}{dt^2}=12t-42}\leftarrow a(t)\\\\
-------------------------------\\\\
s(2)=55\qquad \qquad \qquad \qquad a(2)=-18$

the negative acceleration value, simply means a decreasing rate of change, so, it means the particle is slowing down and possibly coming to a stop before changing direction again.

8 0

3 years ago