Question

Manik wanted to draw a graph of the equation 3x + 7y= 42 on a graph paper, with each square on the graph paper having a side of length one unit. How many of the unit squares in the graph paper will have their interiors lying entirely in the first quadrant and entirely below the graph of the given equations

Answer 1

The unit squares in the graph paper that will have their interiors lying entirely in the first quadrant and entirely below the graph of the given equation is 33

Step-by-step explanation:

The given equation is

$3x+7y=42$

In order to graph this equation, we only need two points

If we put y = 0

We get x = 7

If we put x = 0

We get y = 3

Thus, two points obtained are (7, 0) and (0, 3)

The graph of the equation is attached

The unit squares are counted in the graph

It is clear that the number of unit squares are 33

Hope this answer is helpful.

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Answer 2

Given :   Manik wanted to draw a graph of the equation 3x + 7y= 42 on a graph paper, with each square on the graph paper having a side of length one unit.

To find :  How many of the unit squares in the graph paper will have their interiors lying entirely in the first quadrant and entirely below the graph of the given equations

Solution:

3x + 7y= 42

y = 0   =>  x = 14

y = 1   =>  x  = 35/3  max integer value = 11

Square between y = 0 & 1   =  11

y = 2   =>  x  = 28/3  max integer value = 9

Square between y = 1 & 2   =  9

y = 3   =>  x  = 7  

Square between y = 2 & 3   =  7

y = 4   =>  x  = 14/3  max integer value = 4

Square between y = 3 & 4   =  4

y = 5   =>  x  = 7/3  max integer value = 2

Square between y = 4 & 5   =  2

y = 6   =>  x  =0    

Square between y = 5 & 6   =  0

number of Squares = 11 + 9 + 7 + 4 + 2 + 0

=> 33

There are 33 Squares  having their interiors lying entirely in the first quadrant and entirely below the graph of the given equations

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