Answer:
81.76in^2
Step-by-step explanation:
Given data
Lenght of book= 11.2 inches
WIdth of book= 7.3 inches
We know that Area= Length* Width
Area= 11.2*7.3
Area= 81.76 in^2
Hence the area of the front of the book is is 81.76in^2
Answer: 3
Step-by-step explanation:
4x+23=7x+14
4x-7x=14-23
-3x=9
x=9/3
x=3
In order for the ball to be used in the game, it must be able to meet the minimum and maximum weight requirements. These are the limits of the weight of the ball. If it exceeds the maximum limit of 16 ounces or below the minimum limit of 14 ounces, the ball will not be approved.
So, by adding 1.5 ounces, that would mean that the initial weight of the ball did not reach the minimum limit. The initial weight of the ball, denoted as x, have two possible values. The first value is the initial weight plus added with 1.5 ounces would reach 14 ounces.
x + 1.5 = 14
x = 12.5 ounces
The other scenario is when the initial weight is added to reach the maximum requirement of 16 ounces
x + 1.5 = 16
x = 14.5 ounces
From both answer, we could conclude that the initial weight has to be 12.5 ounces. If the initial weight were 14.5 ounces to begin with, there should be no need for air. It cans till be approved to be used.
Answer:
120 deg
Step-by-step explanation:
Draw a horizontal line that passes vertices of 2 angles 125 deg and 115 deg.
This line is perpendicular with 2 vectors.
Apply the property of vertical angles and the sum of 3 angles in a triangle.
x = 180 - (125 - 90) - (115 - 90) = 120 deg
Answer:
c
Step-by-step explanation:
