Write an inequality that would include the point (3,78), (0,312), (5,10) to be part of the solution.

We want to write an inequality with the points (3, 78), (0, 312) and (5, 10) as solutions.

Then the set of the x-values is {0, 3, 5}

The set of the y-values is {10. 78. 312}

So if we wrote an inequality like:

y > 5

It would include all these solutions, because:

10 > 5

78 >  5

312 > 5

And this has no restriction on the x-variable, so all the x-value range can be a solution, meaning that the points (3, 78), (0, 312) and (5, 10) are solutions to that inequality.

Another one (also trivial) can be:

x < 9

because:

3 < 9

0 < 9

5 < 9

And we have no restriction on the y-values, then the points (3, 78), (0, 312) and (5, 10) are solutions to that inequality.

answer: 7270

step-by-step explanation:

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.