X+ 3 A student solves the equation
2
Il
3x + 5
5 using the steps in the table.
Original equation
x + 3
2
3x + 5
5
Cross multiplication
Distributive property
Subtraction property of equality
5(x + 3) = 2(3x + 5)
5x + 15 = 2(3x + 5)
5 = x
Which method of solving for the variable could be used instead of cross multiplication?
distributing x + 3 and then 3x + 5 to both sides of the equation
distributing x - 3 and then 3x - 5 to both sides of the equation
using the multiplication property of equality to multiply both sides of the equation by 10
using the multiplication property of equality to multiply both sides of the equation by 10

The given equation is:

x+3/2 = 3x+5/5

Cross multiplication would have given us

5(x+3)=2(3x+5)(Equation 1)

Now, if we use the multiplication property of equality to multiply both sides of the equation by 10, we will get:

x+3/2x10=3x+5/5x10

This will become  5(x+3)=2(3x+5), which is the same as (Equation 1) which we had got from cross multiplication.

Thus, out of the given options, the third option, "using the multiplication property of equality to multiply both sides of the equation by 10" is the correct one.

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!

answer is : muchas/tipicas

have a nice day!

~recklesschicken