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3 years ago

13

Plzzz help me plz Show how to find the x-intercepts of this parabola using factoring and the Zero Product Property. Select the a

nswer which is the sum of all solutions to this equation, rounded to the nearest tenth.
y = -10x2 + 8x + 24

1 answer:

DENIUS [597]3 years ago

5 0

Answer:

x = -6/5 and x = 2

Step-by-step explanation:

y = -10x² + 8x + 24

0 = -2(5x+6)(x-2)

5x+6 = 0 x-2 = 0

5x = -6 x = 2

x = -6/5

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How many solutions does the equation x_1 +x_2+x_3+x_4+x_5=21 have where x_1, x_2, x_3, x_4, and x_5 are nonnegative integers and

omeli [17]

Step-by-step explanation:

$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 21\\$ (given)

Let us consider :

$x_{1}$ = $t_{1} + 1$

$x_{2}$ = $t_{2}$

$x_{3}$ = $t_{3}$

$x_{4}$ = $t_{4}$

$x_{5}$ = $t_{5}$

Now, by substituting the above considerations in the above equation, we get:

$t_{1} + 1 + t_{2} + t_{3} + t_{4} + t_{5} = 21\\$

$t_{1} + t_{2} + t_{3} + t_{4} + t_{5} = 20\\$

where,

$t_{i}$ $\geq$ 1

then it follows

n = 20

r = 4

then no. of solutions for the eqn = $_{r}^{n + r}\textrm{C}$

= $_{4}^{24}\textrm{C}$

= 10626

Answer :

no. of solutions for the eqn 10626

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3 years ago

I need help!!!! Someone help please

yarga [219]

Answer:

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3 years ago

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Step-by-step explanation:

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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

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

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3 years ago

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GalinKa [24]

Answer:

B. is the correct answer

4 0

2 years ago