N? + 17n +66 n + 11 n +6

Mathematics

1 answer:

aksik [14]3 years ago

4 0

Answer:

n+11

n+6

Step-by-step explanation:

Assuming you want to select from the given options, the factors of $n^2+17n+66$

We first split the middle term to get:

$n^2+11n+6n+66$

We factor by grouping to obtain:

$n(n+11)+6(n+11)$

We factor further to get:

$(n+11)(n+6)$

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manuel deposits $10000 for 12 yr in an account paying 4% compounded annually.He then puts this total amount on deposit in anothe

Naddik [55]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$10000\\
r=rate\to 4\%\to \frac{4}{100}\to &0.04\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &12
\end{cases}
\\\\\\
A=10000\left(1+\frac{0.04}{1}\right)^{1\cdot 12}\implies A=1000(1.04)^{12}\\\\\\ A\approx 16010.32$

he then turns around and grabs that money and sticks it for another 9 years,

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
~~
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$16010.32\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2\\
t=years\to &9
\end{cases}
\\\\\\
A=16010.32\left(1+\frac{0.05}{2}\right)^{2\cdot 9}\implies A=16010.32(1.025)^{18}
\\\\\\
A\approx 24970.64$

add both amounts, and that's how much is for the whole 21 years.

6 0

3 years ago

Please help right away

amid [387]

Answer:

the second one it shows it going up

8 0

3 years ago

Simplify. Assume that no denominator is equal to zero.

aivan3 [116]

Answer: FIRST OPTION.

Step-by-step explanation:

For this exercise it is important to remember a property called "Quotient of Powers property".

By definition the Quotient of Powers property states the following:

$\frac{a^m}{a^n}=a^{(m-n)}$

You can observe the division of $a^m$ and $a^n$ . Since both have the same base "a" , you must rewrite the base and subtract the exponents "m" and "n".