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3 years ago

If 52/x is a positive integer, how many integer values are possible for x?

Mathematics

1 answer:

vekshin13 years ago

4 0

Since there are 6 factors of 52 (1, 2, 4, 13, 26, 52), the answer is 6

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-. In Ms. Nice's class of 9 students, the average

frutty [35]

Answer: 94.5

Step-by-step explanation:

If average of 9 students is 94, their total will be 94x 9 = 846

If the 10th student scores 99, the new total is 846 + 99 = 945

So, the new average will be

945/10 = 94.5

6 0

3 years ago

Complete the equationof the line whose slope is -2 and y-intercept is (0,-8)

zmey [24]

Answer:hope I ain’t latee

Step-by-step explanation:

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3 years ago

Lisa ran 1/2 of a mile John R and 242 Mi what's girl ran further

julia-pushkina [17]

The fraction that has been given illustrates that the person who ran further is Jane.

<h3>How to solve fraction</h3>

Your information isn't complete. Therefore, an overview of the fraction will given.

Let's assume that Lisa ran 1/2 of a mile and Jane ran 3/4 of a mile. In order to know who ran more, you can convert the fraction to percentage.

This will be:

Lisa = 1/2 × 100 = 50%

Jane = 3/4 × 100 = 75%

Therefore, Jane ran more.

Learn more about fractions on:

brainly.com/question/78672

4 0

2 years ago

The measure of each angle of a regular octagon is _____ the measure of each exterior angle of a regular hexagon.

VikaD [51]

The answer is "A. greater than" because if the number of edges increase, the measure of each exterior angle increases too!

Hope this helps. :)

6 0

3 years ago

Read 2 more answers

What is the completely factored form of f(x)=x^3-7x^2+2x+4

xxMikexx [17]

$Solution, \mathrm{Factor}\:x^3-7x^2+2x+4:\quad \left(x-1\right)\left(x^2-6x-4\right)$

$Steps:$

$x^3-7x^2+2x+4$

$Use\:the\:rational\:root\:theorem,\\a_0=4,\:\quad a_n=1,\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1},\\\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-1,\\\left(x-1\right)\frac{x^3-7x^2+2x+4}{x-1}$

$\frac{x^3-7x^2+2x+4}{x-1}$

$\mathrm{Divide}\:\frac{x^3-7x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3-7x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{x^3}{x},\\\mathrm{Quotient}=x^2,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}x^2:\:x^3-x^2,\\\mathrm{Subtract\:}x^3-x^2\mathrm{\:from\:}x^3-7x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-6x^2+2x+4,\\Therefore,\\\frac{x^3-7x^2+2x+4}{x-1}=x^2+\frac{-6x^2+2x+4}{x-1}$

$\mathrm{Divide}\:\frac{-6x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-6x^2}{x}=-6x,\\\mathrm{Quotient}=-6x,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-6x:\:-6x^2+6x,\\\mathrm{Subtract\:}-6x^2+6x\mathrm{\:from\:}-6x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-4x+4,\\\mathrm{Therefore},\\\frac{-6x^2+2x+4}{x-1}=-6x+\frac{-4x+4}{x-1}$

$\mathrm{Divide}\:\frac{-4x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-4x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-4x}{x}=-4,\\\mathrm{Quotient}=-4,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-4:\:-4x+4,\\\mathrm{Subtract\:}-4x+4\mathrm{\:from\:}-4x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=0,\\\mathrm{Therefore},\\\frac{-4x+4}{x-1}=-4$

$\mathrm{The\:Correct\:Answer\:is\:\left(x-1\right)\left(x^2-6x-4\right)}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

8 0

3 years ago