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netineya [11]
3 years ago
5

Your mother takes u to your grandparents house for dinner. She drives 60 mins at a constant speed of 40 mph. She reaches the hig

hway and quickly speeds up and drives another 30 mins at a constant speed of 70 mph
Tell me how to do this In formula I need to find distance
Mathematics
2 answers:
vlabodo [156]3 years ago
8 0
Distance formula distance=speed×time
Mkey [24]3 years ago
5 0
60 mins= 1 h . 30 mins = 0.5 (1/2) h
before she speeded up, she had gone: s = t×v = 1h ×40 mph = 40 (m)
the length of the high way she drove: s = t×v = 0.5h × 70 mph = 35 (m)
the length of the way from your house to grandparents' house: 40+35=75 (m)
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Common factors of 20w and 40wz
Alla [95]
I hope this helps you

6 0
3 years ago
(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 <em>µ(x, y)</em> 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 <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

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}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
Question in the picture
Mrac [35]
3,3,2 or 5,5,3 Basically just 2 same numbers for the 2 sides of the triangle and 1 other number for the base
3 0
3 years ago
Read 2 more answers
Graph the circle x^2+y^2+2x+4y-44=0
Natasha_Volkova [10]

Answer:

See explanation

Step-by-step explanation:

First, convert this to standard form. By completing the square, you can do the following:

(x^2+2x+1)-1+(y^2+4y+4)-4-44=0

(x+1)^2+(y+2)^2=49

This is a circle with center (-1,-2), and radius 7. It looks like the one I graphed below. Hope this helps!

6 0
3 years ago
What is the value of x in the equation ex- y = 30, when y = 15?
Bumek [7]

Answer:

x = 45

Step-by-step explanation:

Solution:

x - y = 30

x - 15 = 30   ( y = 15 given)

x = 30 + 15

Therefore, x = 45

PLEASE MARK ME AS BRAINLIEST!!!

ALSO DON'T FORGET TO CHECK OUT THIS BLOG WHICH TEACHES YOU SMALL SCIENCE CONCEPTS EVERYDAY AND DONT WORRY ITS ABSOLUTELY FREE AND ITS REALLY BRIEF CONCEPTS:

ITS CALLED dynamicscience

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3 years ago
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