Mohan is selling cookies at the economics fair. As he decides how to package the cookies, he finds that when he bags them in groups of 4, he has 3 left over. When he bags them in groups of 5, he has 2 left over. When he bags them in groups of 7, he has 4 left over. What is the least number of cookies that Mohan could have

67 cookies

Step-by-step explanation:

When packaging in groups of 4, the possible number of cookies is x = (4n+3). The possible values are:

7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71...

When packaging in groups of 5, the possible number of cookies is x = (5n+2). The possible values are:

7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77...

When packaging in groups of 7, the possible number of cookies is x = (7n+4). The possible values are:

11, 18, 25, 32, 39, 46, 53, 60, 67, 74, 81, 88, 95...

The least value that appears on all three possibilities is 67, therefore the least number of cookies that Mohan could have is 67.

ummm seems that u didnt post the attachement !

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c