What is the area of a parallelogram whose vertices are a(−1, 12) , b(13, 12) , c(2, −5) , and d(−12, −5) ?

The area of the parallelogram with the given vertices=238 square units.

Step-by-step explanation:

Let CD be the base of the parallelogram ABCD is given by distance formula

CD=units[or you can count distance by using graph as ordinates of C and D are same to we can count the distance of x from C to D]

Now consider P a point on CD such that AP become the height of the parallelogram which is given by (from the graph attached)=17 [as abscissa of P and A are same to we can count the distance of y from A to P]

Area of parallelogram = base × height=14×17=238 square units.

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !

320ft

step-by-step explanation:

120 divided by 3 equals 40

40 times 4=160

160 times 2=320