Sorry I'm late! but I looked this up on Tiger Algebra and it said "No solution". Hope that helps.
Okay so to find the area you need to find the area of both shapes, then add them to find the final area.
For the square, A=bh
A=(7)(7)
A=49 cm^2
For the triangle, A=1/2bh
A=1/2(3)(7)
A=10.5 cm^2
Add together,
49+10.5 = 59.5 cm^2
I hope that helps!
x + y= 9 and x - y= 3
rearrange the second equation:
x - y= 3
x = 3 + y
x -3 = y
substitute y with "x-3" in the first equation and solve
x + y= 9
x + (x-3)= 9
2x -3 = 9
2x = 12
x = 6
you could also start with the first equation and substitute y in the second.
graphically speaking, it's the intersection point of to lines, at least the six-part of the point.
V = IR...divide both sides by I
V/I = R
583 to the nearest 10 is 580. First you got to find which number is in the tens place. Well in this case 8 is in the tens place. You then look at the number next to it which is 3. 3 is less than 5 so it won't round up it will round down, to a 0.
So..583 to the nearest 10 is 580.