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Nikitich [7]
3 years ago
13

Eleven years ago, you deposited $3,200 into an account. seven years ago, you added an additional $1,000 to this account. you ear

ned 9.2 percent, compounded annually, for the first 4 years and 5.5 percent, compounded annually, for the last 7 years. how much money do you have in your account today? $8,666.67 $7,411.90 $7,717.29 $8,073.91 $8,708.15
Mathematics
1 answer:
vlabodo [156]3 years ago
5 0
$4,550.31 after the first four years before the additional $1,000, $5,550.31 after. $<span>8,073.92 after the last 7 years.</span>
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I need help on this please
yulyashka [42]

Answer:

C

Step-by-step explanation:

The equation of C is 4th-degree, so is definitely non-linear.

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The equation of E can be multiplied by y to get x +y = 2y, which is a linear equation. It graphs as a straight line, so can be called linear on that basis. However, it has a "hole" at y=0, so could arguably be called non-linear.

3 0
4 years ago
Simplify x + 6.2 + 8.5.<br><br> 14.7x<br> x + 14.7<br> 6.2x + 8.5
In-s [12.5K]
Combine the like terms.
Like terms are terms that are the same disregarding the coefficients,

In this case, add the constants.
6.2+8.5=14.7

Final answer: x+14.7
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3 years ago
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In an algebra 2 test and I'm stumped
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7/3-<span>1.41
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aleksandr82 [10.1K]

It's unclear whether the ODE is

xy'+y=4xy^2

or

xy+y'=4xy^2

If the first case, then divide through both sides by y^2 to get

xy^{-2}y'+y^{-1}=4x

then substitute z=y^{-1}, so that z'=-y^{-2}y'. Then

-xz'+z=4x\implies-\dfrac1xz'+\dfrac1{x^2}z=\left(-\dfrac1xz\right)'=\dfrac4x

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\implies z=-4x\ln|x|+Cx

\implies\boxed{y=\dfrac1{Cx-4x\ln|x|}}

If the second, then dividing through by y^2 gives

xy^{-1}+y^{-2}y'=4x

The same substitution as before gives

xz-z'=4x\implies e^{-x^2/2}z'-xe^{-x^2/2}z=\left(e^{-x^2/2}z\right)'=-4xe^{-x^2/2}

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\implies z=4+Ce^{x^2/2}

\implies\boxed{y=\dfrac1{4+Ce^{x^2/2}}}

5 0
3 years ago
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