Describe how to following expression are obtained:\[7xy + 5,{x^2}y,4{x^2} - 5x\]
Answer
661.5k+ views
Hint: An algebraic is the combination of constants and variables. We are the operation like addition subtraction etc. To form an algebraic expression
Variable: - A variable does not have a fixed value. It can be varied. It is represented by letters like a, y, p, m, etc
Constant: - A constant has fix value. Any no. Without a variable is constant.
Terms: - To farm an expression we use constants and variables and separate them using the operations like addition subtraction etc. These parts of expression which we separate using operations are called terms.
Ex: \[4x - y + 7\]
In the above expression, there are three terms, \[4x,y\,and7\]
Factors of a term: -
Every term is the product of its factors, as in the above expression, the term \[4x\] is the product of y and x. So, 4 and x are the factors of that term.
Therefore
Complete step by step answer:
Given
\[7xy + 5..........(1)\]
\[{x^2}y...................(2)\]
\[4{x^2} - 5x...................(3)\]
For equation (1)
\[7xy\] is written as
\[7 \times x \times y\]
\[7xy\] is obtained by multiplying \[7,x\,and\,y\]
Now,
\[7xy + 5\]
The above expression is obtained by adding \[7xy\] with 5.
So, we get \[7xy + 5\]
(ii) Now, for \[{x^2}y\]
\[{x^2} = x \times x\]
\[{x^2}\] is obtained by multiplying \[x\] and \[x\]
\[{x^2}y = {x^2}y\] is obtained by multiplying \[{x^2}\]and y
(iii) \[4{x^2} - 5x\]
\[4{x^2}\] is obtained by multiplying \[4,x\] and \[x\]
And Now,
\[4{x^2} - 5x\]
The above expression is obtained by subtracting the product of 5 and x from the product of \[4,x\] and \[x\]
Note:
Formula and rules using algebraic expression\[ \to \]
The perimeter of an equilateral triangle \[ = 3l\]
\[l = length\]
Perimeter of a square \[ = 4l\]
Area formulas
The area of the square \[ = {a^2}\]
A= side of the square
The area of the rectangle \[ = l \times b\]
Where the length of a rectangle is l and breadth is b
The area of the triangle \[ = \dfrac{1}{2} \times b \times h\]
Where b is the base and h is the height of the triangle.
Here if we know the value of the variables given in the formulas then we can easily calculate the value of the quantity
Variable: - A variable does not have a fixed value. It can be varied. It is represented by letters like a, y, p, m, etc
Constant: - A constant has fix value. Any no. Without a variable is constant.
Terms: - To farm an expression we use constants and variables and separate them using the operations like addition subtraction etc. These parts of expression which we separate using operations are called terms.
Ex: \[4x - y + 7\]
In the above expression, there are three terms, \[4x,y\,and7\]
Factors of a term: -
Every term is the product of its factors, as in the above expression, the term \[4x\] is the product of y and x. So, 4 and x are the factors of that term.
Therefore
Complete step by step answer:
Given
\[7xy + 5..........(1)\]
\[{x^2}y...................(2)\]
\[4{x^2} - 5x...................(3)\]
For equation (1)
\[7xy\] is written as
\[7 \times x \times y\]
\[7xy\] is obtained by multiplying \[7,x\,and\,y\]
Now,
\[7xy + 5\]
The above expression is obtained by adding \[7xy\] with 5.
So, we get \[7xy + 5\]
(ii) Now, for \[{x^2}y\]
\[{x^2} = x \times x\]
\[{x^2}\] is obtained by multiplying \[x\] and \[x\]
\[{x^2}y = {x^2}y\] is obtained by multiplying \[{x^2}\]and y
(iii) \[4{x^2} - 5x\]
\[4{x^2}\] is obtained by multiplying \[4,x\] and \[x\]
And Now,
\[4{x^2} - 5x\]
The above expression is obtained by subtracting the product of 5 and x from the product of \[4,x\] and \[x\]
Note:
Formula and rules using algebraic expression\[ \to \]
The perimeter of an equilateral triangle \[ = 3l\]
\[l = length\]
Perimeter of a square \[ = 4l\]
Area formulas
The area of the square \[ = {a^2}\]
A= side of the square
The area of the rectangle \[ = l \times b\]
Where the length of a rectangle is l and breadth is b
The area of the triangle \[ = \dfrac{1}{2} \times b \times h\]
Where b is the base and h is the height of the triangle.
Here if we know the value of the variables given in the formulas then we can easily calculate the value of the quantity
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