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aleksklad [387]
3 years ago
13

0.008 is 1 tenth of what decimal

Mathematics
1 answer:
shepuryov [24]3 years ago
3 0
0.08, Hope it helped
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A radio station is giving away prizes. They give away CD every 18 minutes and a gift card every 24 minutes. The radio station ga
emmainna [20.7K]

Answer:

Apne see kar Bsdk. Is Bais bano aur desh ko sangali warna patak ke chod denge XD

4 0
3 years ago
What is 0.5, 3/16, 0.75, and 5/45 from least to greatest
BartSMP [9]

5/45, 3/16, 0.5, & 0.75

8 0
3 years ago
Part 2!
olga2289 [7]
Math is about number sense and matching patterns. You get number sense by playing with numbers: counting, arranging. Most people do not find it difficult to match patterns, as our brains are wired to see them--sometimes even when they aren't there.

The picture below is pretty much all you need to know to answer these questions. It is a short course on arithmetic and geometric sequences. (We have left out some simplifications that can be made to the formulas for Nth term, and we have not shown formulas for the sum of a sequence.)

1. Geometric sequence with first term 5, common ratio 2. It will have the formula
.. f(n) = 5*2^(n-1) . . . . . . I can't read your picture to tell whether that is B or D

2. Arithmetic sequence with first term 1 and common difference 3. It will have the formula
.. f(n) = 1 +3(n -1) . . . . . . matches A

3. Geometric sequence with first term 2 and common ratio 3/2 = 1.5. It will have the formula
.. f(n) = 2*1.5^(n -1) . . . . . matches C

4. Geometric sequence with first term 600 and common ratio 60/600 = 0.1. It will have the formula
.. f(n) = 600*0.1^(n-1)
This can be simplified using the rules of exponents to
.. f(n) = 600*(0.1^-1)*(0.1^n)
.. f(n) = 6000*0.1^n . . . . . . matches A

5. The sequence is 12, 11, 10, 9, .... It is an arithmetic sequence with first term 12 and common difference -1. It will have the formula
.. f(n) = 12 +(-1)*(n -1)
.. f(n) = 12 -(n -1) . . . . . . matches B

6. This is an arithmetic sequence with first term 8 and common difference -0.5. It will have the formula
.. f(n) = 8 -0.5(n -1)
.. f(n) = 8.5 -0.5n . . . . . . matches C

7 0
3 years ago
Read 2 more answers
Solve the system of equations x+y=-8 and -9x-6y=60 by combining the equations
TiliK225 [7]

Answer:

x = - 4    y = - 4

Step-by-step explanation:

x+y= - 8

-9x-6y=60

First, solve for x in the first equation:

x+y = - 8       Subtract y from both sides

x + y - y = -8 - y    y cancels on the left

x = - 8 - y

Now plug in what you found for x into the 2nd equation and solve for y.

- 9x - 6y = 60

-9(- 8 - y) - 6y = 60   Multiply out

72 + 9y - 6y = 60

72 + 3y = 60  Subtract 72 from both sides

72 - 72 + 3y = 60 - 72    72 cancels on the left

3y = - 12     Divie both sides by 3

3y/3 = -12/3     3 cancels on the left because 3/3 = 1

y = -4

Now plug your answer for y back into the first equation to get x.

x + y = -8

x + (-4) = - 8  Add 4 to each side

x - 4 + 4 = - 8 + 4   4 cancels on the left

x = -4

x = - 4 and y = - 4

8 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
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