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3 years ago

Which property explains why these two expressions are equal?

Mathematics

1 answer:

STALIN [3.7K]3 years ago

5 0

The correct answer is letter S

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1) Find the lump that must be deposited today to have a future value of $ 25,000 in 5 years if funds earn 6 % componded annually

Hoochie [10]

Answer: $ 18681.45

Step-by-step explanation:

Given: Future value : $FV=\$25,000$

The rate of interest : $r=0.06$

The number of time period : $t=5$

The formula to calculate the future value is given by :-

$\text{Future value}=P(1+i)^n$ , where P is the initial amount deposited.

$\Rightarrow\ 25000=P(1+0.06)^5\\\\\Rightarrow\ P=\dfrac{25000}{(1.06)^5}\\\\\Rightarrow\ P=18681.4543217\approx=18681.45$

Hence, the lump that must be deposit today : $ 18681.45

4 0

3 years ago

−838−1016= ? i dont understand

Maslowich

Answer:

The answer is 236

6 0

3 years ago

Read 2 more answers

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago

What is the sum of this infinite geometric series

nordsb [41]

You probably know that for $|r|$ , we have

$\displaystyle\sum_{i=0}^\infty ar^i=\dfrac a{1-r}$

Your sum starts at $i=1$ , so it's equivalent to

$\displaystyle\sum_{i=1}^\infty 12\left(\dfrac12\right)^i=\sum_{i=0}^\infty 12\left(\dfrac12\right)^i-12\left(\dfrac12\right)^0=\dfrac{12}{1-\frac12}-12=12$

so the answer is A.

4 0

3 years ago

Maggie says if the expression was 12 divided by 3 to the 2nd power you would divide before simplifying is she right?

azamat

Answer:

No she is wrong 3 to the second power first

Step-by-step explanation:

5 0

3 years ago