The length of a rectangular picture frame is twice the width. the frame has a perimeter of 96 inches. write and
solve a system of equations to represent this solution. let x represent the width of the frame, and let y represent the
length of the frame. interpret the solution.

Length = 16 inches

Width = 32 inches

Step-by-step explanation:

Width of the frame is represented by x and length by y inches. It is given that length of the frame is twice the width. This means y is twice of x. So, we can write the equation as:

y = 2x                                      Equation 1

Perimeter of a rectangle is defined as:

Perimeter =  2 ( Length + Width )

Since, perimeter of the picture frame is equal to 96, we can write the equation as:

2(x + y) = 96                          Equation 2

Substituting the value of y from Equation 1 in Equation 2, we get:

2(x + 2x) = 96

2(3x) = 96

6x = 96

x = 16

Substituting the value of x in Equation 1, we get:

y = 2x = 2(16) = 32

This means, the length of the rectangular picture frame is 16 inches and its width is 32 inches.

there will be 35 boys in each team. : )

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.