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3 years ago

5

Solve these compound interest problems and round your answer to the nearest 100th.

1 answer:

MatroZZZ [7]3 years ago

4 0

Answer:

$2,226.96

Step-by-step explanation:

You are going to want to use the compound interest formula, which is shown below.

$A=P(1+\frac{r}{n} )^{nt}$

<em>P = initial balance </em>

<em>r = interest rate </em>

<em>n = number of times compounded annually </em>

<em>t = time </em>

<em />

First, change 10% into its decimal form:

10% -> $\frac{10}{100}$ -> 0.1

Now lets plug in the values into the equation:

$A=500(1+\frac{0.1}{12})^{15(12)}$

$A=2,226.96$

The final amount after 15 years is $2,226.96

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The value of 8 in 5.18 of the value of 8 in 5.81

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Answer:

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3 years ago

Find the points on the lemniscate where the tangent is horizontal. 8(x2 + y2)2 = 81(x2 − y2)

slamgirl [31]

<span>The answers to this problem are:<span>(<span>±5</span></span>√3/8,±5/8)<span>Here is the solution:
Step 1: <span><span><span>x2</span>+<span>y2</span>=<span>2516</span>[2]</span><span><span>x2</span>+<span>y2</span>=<span>2516</span>[2]</span></span>

Step 2: Substitute:<span>
</span><span><span>8<span><span>(<span>25/16</span>)^</span>2</span>=25(<span>x^2</span>−<span>y^2</span>)
</span><span>8<span><span>(<span>25/16</span>)^</span>2</span>=25(<span>x^2</span>−<span>y^2</span>)</span></span>
</span><span>x^2</span>−<span>y^2</span>=<span>25/32</span><span>.

Add [2] and [3]:<span>
</span><span>2<span>x^2</span>=<span>75/32
</span><span>x^2</span>=<span>75/74</span></span>
<span>x=±5</span></span>√3/8<span>
Substitute into [2]:<span>
</span><span><span>75/64</span>+<span>y^2</span>=<span>50/32
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<span>y=±<span>5/8</span></span>

</span>
</span>

3 0

4 years ago

Read 2 more answers

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Triss [41]

Answer:

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Step-by-step explanation:

I can't see anything attached but I know all angles of a triangle add up to 180 degrees. It probably doesn't help tho.

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3 years ago

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