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3 years ago

The first one will get the brianlist please help answer these two

Mathematics

2 answers:

Julli [10]3 years ago

8 0

Answer:

Hello bakugo here!

Step-by-step explanation:

I don't see anything, is there supposed to be anything here? if not

Thank you extra for FREE points! :D

viktelen [127]3 years ago

5 0

Yeah I also don’t see anything

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Subtract. 1 2/5−(−3/5) Enter your answer, in simplest form, in the box.

mars1129 [50]

Answer:

3/5

Step-by-step explanation:

you need to subtract the two fractions then you can divide them by 2 to make them in their simplest form!

HOPE THIS HELPS!!!

5 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 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

3 years ago

Dubnium-262 has a half-life of 34 s. How long will it take for 500.0 grams to

dmitriy555 [2]

Answer:

the time taken for the radioactive element to decay to 1 g is 304.8 s.

Step-by-step explanation:

Given;

half-life of the given Dubnium = 34 s

initial mass of the given Dubnium, m₀ = 500 grams

final mass of the element, mf = 1 g

The time taken for the radioactive element to decay to its final mass is calculated as follows;

$1 = 500 (0.5)^{\frac{t}{34}} \\\\\frac{1}{500} = (0.5)^{\frac{t}{34}}\\\\log(\frac{1}{500}) = log [(0.5)^{\frac{t}{34}}]\\\\log(\frac{1}{500}) = \frac{t}{34} log(0.5)\\\\-2.699 = \frac{t}{34} (-0.301)\\\\t = \frac{2.699 \times 34}{0.301} \\\\t = 304.8 \ s$

Therefore, the time taken for the radioactive element to decay to 1 g is 304.8 s.

4 0

3 years ago

Please help with this problem

Natali [406]

I’m pretty sure it’s chad

6 0

3 years ago

Please help will give brainliest

Gelneren [198K]

Answer:

Step-by-step explanation:

It is B because to find the volume you multiply all three values together, which gets you 68.448. I multiplied 9.2x3.1, which got me 28.52, and then I multiplied 28.52 by 2.4 which got me 68.448.

6 0

3 years ago