WILL GIVE BRAINLIEST!!!!

Mathematics

1 answer:

aniked [119]3 years ago

4 0

It’s 57.66. 50.54+3.8+3.32

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future value of 10% savings from earnings of 36000 earns 6.25% annual interest compounded quarterly for 15 years

hoa [83]

without further ado

10% of 36000 is simply 3600, we chopped a "0" off of it hmmm, ok so

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$3600\\ r=rate\to 6.25\%\to \frac{6.25}{100}\dotfill &0.0625\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years\dotfill &15 \end{cases} \\\\\\ A=3600\left(1+\frac{0.0625}{4}\right)^{4\cdot 15}\implies A=3600(1.015625)^{60}\implies A\approx 9126.53$

7 0

3 years ago

Let <img src="https://tex.z-dn.net/?f=%5Calpha" id="TexFormula1" title="\alpha" alt="\alpha" align="absmiddle" class="latex-form

Margarita [4]

Answer:

$|\alpha| = 2$

Step-by-step explanation:

Since $\alpha$ and $\beta$ are complex conjugates, let's define them as follows:

$\alpha = a+bi$

$\beta = a-bi$

$\frac{\alpha}{\beta^2}=\frac{a+bi}{a^2-b^2-2abi} =\frac{(a+bi)*(a^2-b^2+2abi)}{(a^2-b^2-2abi)*(a^2-b^2+2abi)} =\frac{a^3-3ab^2+(3a^2b-b^3)i}{a^4+2a^2b^2+b^4}$

Since $\frac{\alpha}{\beta^2}$ is a real number, complex part of above result must be zero.

$3a^2b-b^3=0$

From to hold above equality, $b=0$ or $b^2=3a^2$ .

However, since $|\alpha-\beta|=2\sqrt 3$ , $b\neq 0$

So, $b =\sqrt 3 a$ or $b =-\sqrt 3 a$

And since $\alpha$ and $\beta$ are complex conjugates, taking plus or minus sign as found above will not affect the result, so let's write the last version of $\alpha$ and $\beta$ as follows:

$\alpha = a+\sqrt 3 ai$