The perimeter of kite wxyz is 2√53+ __ units.

The perimeter of the kite, using the formula to calculate the distance between two points, is 2√53+10 units

Solution:

Perimeter of kite WXYZ: P=?

The kite is a polygon, and the perimeter of a polygon is the sum of the lengths of its sides. In this case:

P=WX+XY+YZ+WZ

1. WX is an horizontal segment, then its length is the difference of the abscissas of their end points:

WX=Xx-Xw

W=(-3,3)=(Xw,Yw)→Xw=-3, Yw=3

X=(2,3)=(Xx,Yx)→Xx=2, Yx=3

Replacing the values:

WX=Xx-Xw

WX=2-(-3)

WX=2+3

WX=5 units

2. XY is an oblique segment, then to determine its length we must use the formula to calculate the distance between two points:

If P1=(x1,y1) and P2=(x2,y2)→P1P2=√[(x2-x1)^2+(y2-y1)^2]

XY=√[(Xy-Xx)^2+(Yy-Yx)^2]

X=(2,3)=(Xx,Yx)→Xx=2, Yx=3

Y=(4,-4)=(Xy,Yy)→Xy=4, Yy=-4

Replacing the values:

XY=√[(Xy-Xx)^2+(Yy-Yx)^2]

XY=√[(4-2)^2+(-4-3)^2]

XY=√[(2)^2+(-7)^2]

XY=√(4+49)

XY=√53 units

3. YZ is an oblique segment, then to determine its length we must use the formula to calculate the distance between two points:

If P1=(x1,y1) and P2=(x2,y2)→P1P2=√[(x2-x1)^2+(y2-y1)^2]

YZ=√[(Xz-Xy)^2+(Yz-Yy)^2]

Y=(4,-4)=(Xy,Yy)→Xy=4, Yy=-4

Z=(-3,-2)=(Xz,Yz)→Xz=-3, Yz=-2

Replacing the values:

YZ=√[(Xz-Xy)^2+(Yz-Yy)^2]

YZ=√[(-3-4)^2+(-2-(-4))^2]

YZ=√[(-7)^2+(-2+4)^2]

YZ=√[(-7)^2+(2)^2]

YZ=√(49+4)

YZ=√53 units

4. WZ is a vertical segment, then its length is the difference of the ordinates of their end points:

WZ=Yw-Yz

W=(-3,3)=(Xw,Yw)→Xw=-3, Yw=3

Z=(-3,-2)=(Xz,Yz)→Xz=-3, Yz=-2

Replacing the values:

WZ=Yw-Yz

WZ=3-(-2)

WZ=3+2

WZ=5 units

Replacing the values in the formula of perimeter:

P=WX+XY+YZ+WZ

P=5 units+√53 units+√53 units+5 units

Adding similar terms:

P=2√53+10 units

x is greater than 5

step-by-step explanation:

2x+7 is greater than 17

    -7                           -7

2x is greater than 10

divide by 2 each side

x is greater than 5

70 years.

step-by-step explanation: