All categories

3 years ago

What is the are of the square?(look at photo)

Mathematics

2 answers:

grigory [225]3 years ago

7 0

Answer:

2.4 use a scientific calculator like desmos it will really help:)

Step-by-step explanation:

Gemiola [76]3 years ago

4 0

Answer:

0.36 or 9/25

Step-by-step explanation:

3/5*3/5

You might be interested in

Please solve e e eee

quester [9]

Answer:

Step-by-step explanation:

Multiply both sides by 15, which gives you r≥-75

Since the graph has to go in the direction of numbers greater than -75, that would be -74, -73, -72, etc. So, it is C.

8 0

2 years ago

Which inequality is NOT true when x= -2?

Levart [38]

Answer: D

Step-by-step explanation:

-7 isn't greater than -4

3 0

3 years ago

This doesn't tell me anything what is this supposed to mean, can you please explain it!!

KiRa [710]

Normally I'd love to help, but put in a picture. I'm confused.

7 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 :

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