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3 years ago

Choose the correct conic section to fit the equation. X=1/16y^2

Mathematics

1 answer:

arsen [322]3 years ago

3 0

Hello from MrBillDoesMath!

Answer:

Parabola

Discussion:

The equation is x = (1/16) y^2. Multiply both sides by 16 to get

y^2 = 16x is a parabola that "opens to the right". (Looks like y = x^2 rotated clockwise 90 degrees)

Thank you,

MrB

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:Phil’s age decreased by 9 years is 14 years ?

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Answer:

Phil is 23

Step-by-step explanation:

Add 14 to 9 and you get 23. So then you subtract 9 from 23

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2 years ago

What is the mean of 85,97,84,88,95,100,81

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1 year ago

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Given a triangle with side lengths 4.1 and 1.3, what is the range of possible sizes for x

zimovet [89]

Answer:

2.8-5.4

Step-by-step explanation:

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3 years ago

Solve the system using the substitution method<br> 6x - 2y = -4<br> y= 3x + 2

Anit [1.1K]

Answer:

Step-by-step explanation:

6x - 2y = -4

y= 3x + 2

We see that y = 3x + 2 so we can use that value of Y everytime we see i in the other equation.

6x - 2(3x +2) = -4

Now usually we'd we simply solve for X.

6x - 6x -4 = -4

This clearly does not work as we cannot get rid of X

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3 years ago

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Inverse laplace of [(1/s^2)-(48/s^5)]

Katen [24]

**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}$

Given that $\dfrac{1}{s^2} = \dfrac{1!}{s^2}$ and $-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$ and hence $\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}$ .

Hope this helps...

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3 years ago