How do i setup this problem? ? i need some

7 nickels and 10 dimes

Step-by-step explanation:let

n = amount of nickels

d = amount of dimes

n + d=16---(1)

For the second equation, we know that a dime is worth 10¢ and a nickel is 5¢, so it should be:

5n + d =125

(n+2d)=25

n+2d=25(2)

Equation (1) - (2)

n+d-n-2d=16-25

d=9(3)

Substituting value of d from equation (3) to (1)

n+9=16

n=7

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!