The second one, fourth one, and the fifth one
First, write an equation system based on the problem
An equation for "the perimeter of a rectangle is 70 in" can be written as following
∴ 2l + 2w = 70 <em>(first equation)</em>
An equation for "the length is 11 in more than its width" can be written as following
∴ l = w + 11 <em>(second equation)</em>
Second, solve the equation by substitution method o find the dimension of the rectangle.
Substitute/plug l as (w+11) into the first equation
2l + 2w = 70
2(w + 11) + 2w = 70
2w + 22 + 2w = 70
4w + 22 = 70
4w = 70 - 22
4w = 48
w = 48/4
w = 12
The width of the rectangle is 12 in
Substitute w with 12 to the second equation
l = w + 11
l = 12 + 11
l = 23
The length of the rectangle is 23 in
Third, find the area of the rectangle
a = l × w
a = 23 × 12
a = 276
The area of the rectangle is 276 in²
Answer:
C
Step-by-step explanation:
14 + w is the same as w + 14
Answer:
136.015 yd
Step-by-step explanation:
Length of diagonal is:
√( (80 yd)² + (110 yd)² ) = √( 18500 yd² ) = 136.015 yd
The dilation it’s increasing by is 4
If you multiply each of the sides by 4 you’ll get the sides of triangle B