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Bogdan [553]
3 years ago
13

Help with this math question please

Mathematics
2 answers:
Cloud [144]3 years ago
4 0
A = P( 1 + rt)
A/P = 1 + rt
A/P - 1 = rt
t = (( A/P)) - 1)/r


t = (( 5500/1000) - 1)/(6.25/100
t = (5.5 - 1)/0.0625
t = 4.5/0.0625 = 72 years


Answer it will  take 72 Years...



Hope it helps!!!!!
Masja [62]3 years ago
3 0
100
100 \\ \\ \times 55 = 5500
55 weeks 1 year there 52 weeks in one year
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Solution for dy/dx+xsin 2 y=x^3 cos^2y
vichka [17]
Rearrange the ODE as

\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y
\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3

Take u=\tan y, so that \dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}.

Supposing that |y|, we have \tan^{-1}u=y, from which it follows that

\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}
\sec^2y=1+\tan^2y=1+u^2

So we can write the ODE as

\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3

which is linear in u. Multiplying both sides by e^{x^2}, we have

e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}
\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}

Integrate both sides with respect to x:

\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx
e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx

Substitute t=x^2, so that \mathrm dt=2x\,\mathrm dx. Then

\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt

Integrate the right hand side by parts using

f=t\implies\mathrm df=\mathrm dt
\mathrm dg=e^t\,\mathrm dt\implies g=e^t
\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)

You should end up with

e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C
u=\dfrac{x^2-1}2+Ce^{-x^2}
\tan y=\dfrac{x^2-1}2+Ce^{-x^2}

and provided that we restrict |y|, we can write

y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)
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3 years ago
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artcher [175]
Positive infinity and negative infinity 

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3 years ago
National Income (NI) is the sum of the incomes that all individuals in an economy earn in the forms of wages,
Dennis_Churaev [7]

Answer:

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During the period 1994 - 2004, the 'National Income' ,(NI) of Australia grew about 5.2% per year (measured in 2003 U. S, dollars).  In 1994 , the NI of Australia was $ 4 billion.

Now,

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Assuming this rate of growth continues, the NI of Australia in the year 2020 (in billion dollars) will be,

4 \times[\frac{(100 + 5.2)}{100}}]^{26}

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What are the coordinates of the point (−1, 6) after a reflection across the x-axis?
Agata [3.3K]
The answer would be (-1,-6). Remember that the x-axis runs left to right and the y-axis runs up and down. Therefore, if you were to reflect it, just picture it being flipped over. If you go from (-1,+6) and you reflect it over the x-axis, both points are now in the bottom left quadrant where everything is negative. So, you get (-1,-6) The -1 doesn't change since it was already a -x value before and stayed in a negative area.
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3 years ago
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