An intelligent trader travels from 1 place to another carrying 3 sacks having 30 coconuts each. No sack can hold more than 30 coconuts. On the way
he passes through 30 checkpoints and on each checkpoint he has to give 1
coconut for each sack he is carrying. How many coconuts are left in the
end? *

none

Step-by-step explanation:

the man is carrying 3 sacks

each has 30 coconuts

so at total he has 90 coconuts:  30*3= 90  

he passe through 30 checkpoints

ha has to give 1 coconut for each sack

so he gives 3 coconuts each time: 3*1=3

there are 30 ckeckpoints so : 3*30= 90

he has spent all the coconuts unless he has a trick

He starts with 3 sacks with 30 in each sack.

He has to give 1 coconut per sack away. So at first he gives 3 coconuts away.

The first 10 checkpoints he gives away 30 coconuts, so he is left with 2 sacks.

Now he has to give 2 coconuts away. 30/2 = 15, so the next 15 checkpoints he ends up giving away another full sack, so he is left with 1 full sack of 30 coconuts and he has 5 checkpoints left.

Giving away 1 coconut at those checkpoints, he would have 25 left

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .