Given f(x) = -x – 1, solve for a when f(x) = 6.

A credit union client deposits $3,300 in an account earning 6.5% interest, compounded annually. What will the balance of the acc

astra-53 [7]

Answer:

The balance of account at the end of 36 years is $31,849.29

Step-by-step explanation:

We are given the following in the question:

P = $3,300

r = 6.5% = 0.065

t = 36 years

The compound interest is given by:

$A = p\bigg(1+\dfrac{r}{n}\bigg)^{nt}$

where A is the amount, p is the principal, r is the interest rate, t is the time in years and n is the nature of compound interest.

Since interest is compounded annualy we use n = 1

$A = 3300\bigg(1+\dfrac{0.065}{1}\bigg)^{36}\\\\A = \$31,849.29$

Thus, balance of account at the end of 36 years is $31,849.29

6 0

2 years ago

Help, i will give brainliest

andrey2020 [161]

Answer:

The answer is J

Step-by-step explanation:

If you start from Quinn's location (8,7), and move south 3 units, you get to (8,4). Then you move to the west 4 units and that gets you to (4,4).

hope this helped! :)

8 0

3 years ago

What is the true solution In 20+In 5=2 In x?

S_A_V [24]

Answer:

x = 10

Step-by-step explanation:

Using the rules of logarithms

• log x + log y ⇔ log(xy)

• log $x^{n}$ ⇔ n log x

• log x = log y ⇒ x = y

Given

ln20 + ln5 = 2 lnx , then

ln(20 × 5) = ln $x^{2}$

ln 100 = ln $x^{2}$ , hence

x² = 100 ( take the square root of both sides )

x = 10

7 0

3 years ago

A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest

faltersainse [42]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
A=4000\left(1+\frac{0.02}{1}\right)^{1\cdot 4}\implies A=4000(1.02)^4\implies A\approx 4329.73$

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396$

add both amounts, and that's her investment for the 11 years.

7 0

2 years ago

What is the answer? <br> Expand and simplify 6(2x-3)-2(2x+1)

vitfil [10]

Answer:

8x-20

Step-by-step explanation:

Distribute:

=(6)(2x)+(6)(−3)+(−2)(2x)+(−2)(1)

=12x+−18+−4x+−2

Combine Like Terms:

=12x+−18+−4x+−2

=(12x+−4x)+(−18+−2)

=8x+−20

Answer:

=8x−20

7 0

2 years ago