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2 years ago

5

The ratio of white marbles to blue marbles and jays bag is 2 to 5 if there is there are more than 50 marbles in the bag what is

the minimum number of marbles that can be in the bag?

1 answer:

Ilya [14]2 years ago

5 0

Answer:

If it is blue marbles than probably 30. White marbles, then maybe 20.

Step-by-step explanation:

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 <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

Please help! Picture included!!!

V125BC [204]

The answer is c)32 hope this helps

5 0

3 years ago

Read 2 more answers

How many sides does a regular polygon have with each exterior angle measures up to 30

ch4aika [34]

The answer would be 6 exteriors.

5 0

3 years ago

Which expression is equivalent to the expression 4(6+y)+7?

natita [175]

Answer:

31 + 4y

Step-by-step explanation:

4(6+y)+7

Expand the brackets.

24+4y+7

Rearrange.

4y + 24 + 7

Add the like terms.

4y + 31

7 0

3 years ago

Write an algebraic expression that represents each purchase.

Svetradugi [14.3K]

Answer: Both answers will be

$15x+18$

Step-by-step explanation:

Since we have given that

Cost of soccer balls = $15

Cost of baseball = $6

Cost of sweat socks = $5

a) Mr. Tonkery bought x number of soccer balls and 3 base balls .

So, our algebraic expression will be

$15x+3\times 6\\\\=15x+18$

b) Dennis, Eddie and Felix are on a base ball team. They each bought a baseball and x pair of sweat socks .

Since there are 3 members in a base ball team ,

Number of base ball they will purchase is given by

$3\times 6=18$

Number of sweat socks they will bought is given by

$5\times x\times 3=15x$

Total number they will purchase is given by

$15x+18$

Hence, both answers will be

$15x+18$

6 0

3 years ago