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3 years ago

Calculate 3n-1 for n equals 2 to 10 explain how you know that the numbers are composite

Mathematics

1 answer:

Troyanec [42]3 years ago

5 0

composite number means a number that can be formed by multiplying two smaller numbers

3n-1 for n=2, 3n-1=3*2-1=6-1=5

3n-1 for n=10, 3n-1=3*10-1=30-1=29

5 is composite because 2*2.5=5

29 is composite because 5*5.8=29

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What is the volume of the composite figure shown below?

AVprozaik [17]

Answer:

2412 mm³

Step-by-step explanation:

In the figure attached, the composite figure is shown.

Dimension of top rectangular prism:

length of 13 millimeters

width of 12 millimeters

height of 12 millimeters

Volume of top rectangular prism: length*width*height = 13*12*12 = 1872 mm³

Dimension of bottom rectangular prism:

length of 15 millimeters

width of 12 millimeters

height of 3 millimeters

Volume of bottom rectangular prism: length*width*height = 15*12*3 = 540 mm³

Volume of the composite figure: 1872 + 540 = 2412 mm³

7 0

3 years ago

−2(3x−4)−x−3= \,\,33 33

notsponge [240]

Answer: $=-7x+5$

Step-by-step explanation:

$-2\left(3x-4\right)-x-3$

$=-\left(2\left(3x-4\right)+x+3\right)$

$=-\left(6x-8+x+3\right)$

$=-\left(7x-5\right)$

$=-7x+5$

5 0

3 years ago

The two different types of cells are please tell me the answer

vodomira [7]

Answer:

eukaryotic cells (cells that have a nucleus) and prokaryotic cells (cells that do not have a nucleus)

Step-by-step explanation:

5 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

3 years ago

22/110 to its lowest terms

Dmitrij [34]

Find the GCF, in this case 11, and divide both the numerator and denominator by it. Here, it turns out to be 2/11 because 22÷11 is 2 and 110÷11 is 11.

3 0

3 years ago

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