Ask question

Ask question

All categories

4 years ago

10

Currently about 9 in 10 people own a cell phone in the US. If there are 428 people on an airplane, approximately how many of the

m own a cell phone and need to turn it off.
375

380

385

390

1 answer:

lisov135 [29]4 years ago

5 0

A way to solve this is finding 90% of 428.

428 × 0.90 = 385.2
Rounded, the answer is 385

You might be interested in

What is the easiest way to multiplying decimals

bezimeni [28]

Hi so the easiest way you can do it is

Step 1:Multiply normally, ignoring the decimal points put the decimal point in the answer.

Step 2: then put the decimal point in the answer - it will have as many decimal places as the two original numbers combined. Hope that help!

6 0

4 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 <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

Cone because if said 1 circular base and 1 cuved that is a cone

OLEGan [10]

Yes this answer is correct

5 0

3 years ago

Solve sin x = cos x on the closed interval [pi, 3pi/2].

Gwar [14]

Answer: x=5π/4

Step-by-step explanation:

sin x = cos x

sin^2 x=cos^2 x

sin^2 x+sin^2 x=cos^2 x+sin^2x

2 sin^2 x=1

sin^2 x=1/2

sin x=(1/2)^1/2

sin x= $\frac{\sqrt{2}}{2}$

x=45 degrees, 225 degrees

x=180/4, 5(180)/4

x=π/4, 5π/4 ==> π<=x<=3π/2

x=5π/4

5 0

1 year ago

How does 0.03 compare to 0.003

prohojiy [21]

<span>0.03 compare to 0.003
first which one is higher?
0.03 is 10 times higher than 0.003.
to get the answer, multiply 0.003 by 10
=> 0.003 x 10 = 0.03

And the value of 0.03 is hundredths and 0.003 is thousandths.</span>

8 0

3 years ago

Read 2 more answers