All categories

2 years ago

36 is 0.9% of what to answer w

Mathematics

1 answer:

Alexus [3.1K]2 years ago

6 0

Answer: w=4,000

Step-by-step explanation:

Let w=the number, 36 is 0.9% of number w and w represent the 100%

so using proportion we can find out the number by using cross multiplication.

w/100%=36/.9%

w=100*36/.9

w=4,000

Hope this helps, HAVE A BLESSED AND WONDERFUL DAY! As well as a great Superbowl Weekend! :-)

- Cutiepatutie ☺❀❤

You might be interested in

(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 :

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

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 F(x, y) = C. 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 x :

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

Differentiate both sides of this with respect to y :

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

Then the general solution to the DE is

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

5 0

2 years ago

Prove that <img src="https://tex.z-dn.net/?f=cos%209x-cos%205x%2Fsin17x-sin3x%3D-sin2x%2Fcos10x" id="TexFormula1" title="cos 9x-

maria [59]

Cos9x - cos5x = -2sin2x.sin7x
sin17x - sin3x = 2sin7x.cos10x
so f(x) = $\frac{-2sin2x.sin7x}{2sin7x.cos10x}$ = $\frac{-sin2x}{cos10x}$

8 0

2 years ago

Explain how the zero property in the problem 2/3 (0) = 0 (2/3) = 0 is the same with any rational number.

Gala2k [10]

Its because the power of zero is undefined

6 0

2 years ago

Find the geometric mean of the pair of numbers.

tatyana61 [14]

The formula for geometric mean is nth root of the product of all terms. n denotes the number of data. Since you have only two points, 99 and 11, n = 2. The product of both terms is 99 × 11 = 1089. The square root of 1089 is 33. Thus, the answer is 33.

6 0

2 years ago

Read 2 more answers

Can someone explain #1 and #2 to me? (Picture above) I really need help

Charra [1.4K]

A number can replaced with a variable like x
(X+3)4=16
4x+12=16
4x=4
X=1
(X/4)+3=24
X/4=21
X=84

5 0

2 years ago