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3 years ago

Use the quadratic formula to solve the equation. Round to the nearest hundredth if needed.

Mathematics

1 answer:

klasskru [66]3 years ago

8 0

The answer is above.

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Please Help and Thanks.

Katena32 [7]

Answer:

Step-by-step explanation:

case 1

cost of 20oz bottle of soda = $1.20

dividing both side by 20

cost of 20/20 oz bottle of soda = $1.20/20

cost of 1oz bottle of soda = $0.06

__________________________________________

Case 2

cost of 12oz bottle of soda = $0.85

dividing both side by 12

cost of 20/20 oz bottle of soda = $0.85/12

cost of 1oz bottle of soda = $0.07

Based on above calculation, we can see that first deal 20oz bottle of soda at costs $1.20 is better deal as per unit cost is $0.06 which is less than 12 oz can of soda costs $0.85 which is equivalent to $0.07 per unit cost

7 0

2 years ago

Express the perimeter of one side of the square using s

r-ruslan [8.4K]

s is the length and w is width which the equation is 2s+2w=P

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 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

A passenger train has tickets available for 12 window seats and 8 aisle seats. The nest person to buy a ticket will be randomly

nydimaria [60]

The probability of the person be set to sit in an aisle seat is 8/12.

4 0

3 years ago

Read 2 more answers

Find the product of (3x²y²) (4xy³)²

Elena-2011 [213]

First, I would solve the second parenthesis.

(4xy^3)^2

Distribute

(4^2 x^2 y^6)

4 x 4 = 16

Now, combine like terms

3x^2 x 4x^2 = 12x^4

y^2 x y^6 = y^12

So, the answer would be 12x^4 y^12

Hmm actually I'm not sure. I did this about two years ago so I don't really remember sorry if this is really wrong

5 0

2 years ago

Read 2 more answers