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3 years ago

Write the prime factorization for 65,169

Mathematics

1 answer:

posledela3 years ago

8 0

Let's try to find some primes that divide this number.

The number is not divisible by 2, because it is odd.

The number is divisible by 3 though, because the sum of its digits is:

$6+5+1+6+9=27=3\cdot 9$

So, we can divide the number by 3 and keep going with the factorization:

$65169\div 3 = 21723$

This number is again divisible by 3, because

$2+1+7+2+3 = 15 = 3\cdot 5$

We have

$21723\div 3 = 7241$

This number is no longer divisible by 3. Let's go on looking for primes that divide it: 5 doesn't because the number doesn't end in 0 nor 5. This number is not divisible by 7 or 11 either (just try). It is divisible by 13 though: we have

$7241\div 13 = 557$

And 557 is prime, so we're done. This means that the prime factorization of 65169 is

$3^2\cdot 13 \cdot 557$

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NEED HELP! [URGENT] Submit the worksheet with your constructions to your teacher to be graded.

Tresset [83]

give me more details. that's a bit confusing

4 0

3 years ago

I already graphed the first question, need number 2 please

Soloha48 [4]

For the equation y = 2x + 3, calculate the x and y intercepts

when y = 0

0 = 2x + 3

x = -3/2

x = -1.5

The x-intercept = (-1.5, 0)

when x = 0

y = 2(0) + 3

y = 3

The y-intercept = (0, 3)

Locate the points (-1.5, 0) and (0, 3) to graph the equation y = 2x + 3

For the equation y = -1/3 x + 2, calculate the x and y intercepts

when y = 0

$\begin{gathered} 0=-\frac{1}{3}x+2 \\ \frac{1}{3}x=2 \\ x=6 \end{gathered}$

The x-intercept = (6, 0)

when x = 0

y = -1/3 (0) + 2

y = 2

The y-intercept = (0, 2)

The graphs of the two equations are plotted using their x and y intercepts as shown below

y = 2x + 3 is plotted in red

y = -1/3 x + 2 is plotted in blue

The solution to the system of equations represented by the two lines is (-0.429, 2.143)

6 0

1 year ago

Please help I will mark brainlest

spin [16.1K]

Hope this helps, pls mark me brainliest

4 0

3 years ago

Write an equation of cosine function with amplitude 2 and a period 4pi

Arisa [49]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now.. hmmm let's see, keeping in mind the template above

so, amplitude of 2, A=2, thus |A| = ±2, so, you can use either

and period of 4π, well, that simply means that

$\bf \cfrac{2\pi }{B}=4\pi \implies \cfrac{2\pi }{4\pi }=B\implies \cfrac{1}{2}=B\\\\
-----------------------------\\\\

\begin{array}{llll}
f(\theta)=&\pm 2cos&\left(\frac{1}{2}\theta \right)\\
&\ \uparrow &\ \uparrow \\
&A&B
\end{array}$

8 0

4 years ago

Acellus inverse variation

RUDIKE [14]

The current when the resistance is 10 ohms is 24 amps

<h3>What are variations?</h3>

Variations are simply data that change in values (i.e. not constant)

<h3>Types of variation</h3>

The types of variations are: