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2 years ago

Use the following terms to complete the following statements: integers, rational

Mathematics

2 answers:

asambeis [7]2 years ago

8 0

Answer:

the counting numbers and zero are whole numbers.

the counting numbers and their opposites, along with zero, are integers.

integers and decimal equivalents of fractions are rational numbers.

lozanna [386]2 years ago

5 0

Answer:

Answer:

the counting numbers and zero are <u>whole numbers.</u>

the counting numbers and their opposites, along with zero, are<u> integers. </u>

integers and decimal equivalents of fractions are<u> rational numbers.</u> hoped this helped

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A rectangular building for a gym is three times as long as it is wide. Just inside the walls of the building, there is a 6ft rec

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The equation 3x² - 48x + 6856 represents the area of the gym and track together in terms of width.

<h3>What is the area of the rectangle?</h3>

It is defined as the area occupied by the rectangle in two-dimensional planner geometry.

The area of a rectangle can be calculated using the following formula:

Rectangle area = length x width

We have:

A rectangular building for a gym is three times as long as it is wide. Just inside the walls of the building, there is a 6ft rectangular track along the walls of the gym and has an area of 7000ft²

Let x be the width of the rectangle.

As the area of track along the walls of the gym is 7000ft²

(x - 12)(3x - 12) = 7000

After simplifying:

3x² - 48x + 6856 = 0

Thus, the equation 3x² - 48x + 6856 represents the area of the gym and track together in terms of width.

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1 year ago

Twenty less than a number is more than twice the same number . Step by step to get the answer

likoan [24]

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3 years ago

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2 years ago

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What are the number of integral solutions of the equation 2x+2y+z=20 such that x>=0 , y>=0 , z>=0?

zaharov [31]

$2x+2y+z=20\\
z=\dfrac{20}{2x+2y}\\
z=\dfrac{10}{x+y}$

Now, for z to be an integer, the sum x+y must be a divisor of 10.
It has to be a positive divisor since z≥0. Also x≠y.

$x+y=1 \\
x=1-y\\
$
x≥0 and y≥0 so y can be equal to either 0 or 1. There are 2 solution in this case.

$x+y=2\\
x=2-y\\$
In this case, y can be equal to 0,1, or 2, but for y=1 ⇒ x=1, so there are two solutions.

$x+y=5\\
x=5-y$
y can be 0,1,2,3,4 or 5 - 6 solutions

$x+y=10\\
x=10-y$
y can be 0,1,2,3,4,5,6,7,8,9,10, but for y=5 ⇒ x=5, so 10 solutions.

2+2+6+10=20 solutions in total.

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3 years ago