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Olin [163]
3 years ago
11

Write an algebraic expression that is the quotient of a variable term and a constant.

Mathematics
1 answer:
Korolek [52]3 years ago
4 0
<h3><u>Question:</u></h3>

Write an algebraic expression that is the quotient of a variable term and a constant

<h3><u>Answer:</u></h3>

The algebraic expression that is the quotient of a variable term and a constant is quotient = \frac{x}{5}

<h3><u>Solution:</u></h3>

In arithmetic, a quotient is the quantity produced by the division of two numbers

Here given in question that quotient of a variable term and a constant

So the quotient produced by division of variable term and constant

A variable is a special type of amount or quantity with an unknown value. Though it can be anything, you often see letters like x or y as variables in algebraic equations

Let the variable term be "x"

Constants are the terms in the algebraic expression that contain only numbers

Let the constant be "5" (Note that it can be any number, here we choose 5)

So the quotient of a variable term and a constant is:

An algebraic expression is a mathematical expression that consists of variables, numbers and operations.

quotient = \frac{\text{ variable term}}{constant}

quotient = \frac{x}{5}

quotient = x \div 5

<h3><u>Question:</u></h3>

Each side of a square has a length of 5x. Use your area expression to find the area of the square when x = 2.2 centimeters. Show your work.

<h3><u>Answer:</u></h3>

The area of square is 121 square centimeter

Solution:

Given that each side of square has length of 5x

The area of square is given as:

\text{ area of square }= (side)^2

\text{ area of square} = (5x)^2\\\\\text{ area of square} = 25x^2

Given x = 2.2 centimeter

\text{ area of square} = 25(2.2)^2 = 25 \times 4.84 = 121

Thus area of square is 121 square centimeter

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Answer:

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6 0
3 years ago
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DiKsa [7]
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6 0
4 years ago
Solve the equation for y in terms of x, and replace y with function notation f(x). Then find f(2).
stealth61 [152]

Answer:

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Step-by-step explanation:

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7 0
4 years ago
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