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3 years ago

Can someone pls help, I know this is kind of a lot but I’ve been stuck on this for a while now.

Mathematics

1 answer:

Greeley [361]3 years ago

3 0

9514 1404 393

Answer:

(a, b, c) = (-0.425595, 11.7321, 2.16667)

f(x) = -0.425595x² +11.7321x +2.16667

f(1) ≈ 13.5

Step-by-step explanation:

A suitable tool makes short work of this. Most spreadsheets and graphing calculators will do quadratic regression. All you have to do is enter the data and make use of the appropriate built-in functions.

Desmos will do least-squares fitting of almost any function you want to use as a model. It tells you ...

a = -0.425595

b = 11.7321

c = 2.16667

f(x) = -0.425595x² +11.7321x +2.16667

and f(1) ≈ 13.5

_____

Additional comment

Note that a quadratic function doesn't model the data very well if you're trying to extrapolate to times outside the original domain.

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The sum of 3x and 1 is greater than-5 and less than or equal to 10

ira [324]

Answer:

x is greater than -2 and less than or equal to 3

Step-by-step explanation:

so the 3x + 1 is greater than -5

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we also know that 3x + 1 is less than or equal to 10

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2 years ago

Which pairs of rectangles are similar polygons? Select each correct answer.

iragen [17]

Answer:

The bottom two

Step-by-step explanation:

[] Similar beings their sides have equivalent ratios to each other

[] See attached

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

8 0

3 years ago

Jonas weight 63 pounds. jane weight 67 pounds. round each number to the nearest ten.

IrinaVladis [17]

Answer:

Jonas = 60 pounds

Jane = 70 pounds

Remember- 5 or more, raise the score. 4 or less, let it rest.

Jonas is 63 pounds. 3 is less than 4, so rounded to the nearest ten would be 60 pounds.

Meanwhile, Jane is 67 pounds. 7 is more than 5, so we will raise the score.

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4 0

3 years ago

Read 2 more answers

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$