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:
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This was the example answer. Hope this helps!! HAVE AN AMAZING DAY!
1.
a^2+b^2=c^2
easy
a=a
b=13
c=21
a^2+13^2=21^2
a^2+169=441
minus 169 both sides
a^2=272
sqrt both sides
a=16.4924
D is answer
2. same thing
a=18
b=26
18^2+26^2=c^2
324+676=c^2
1000=c^2
sqrt both sides
31.6228=c
C is answer
3.
one way is to plug them in
remember that hypotonuse is longest side
A. 7^2+24^2=25^2, is that true?, yes it is treu
answer is A
4.
legs are 6 and 4
a=6
b=4
c=x
4^2+6^2=x^2
16+36=x^2
52=x^2
sqrt both sides
7.2111=x
7.2=x
5.c=20
a=x
b=14
x^2+14^2=20^2
x^2+196=400
minus 196 both sides
x^2=204
sqrt both sides
x=14.2829
x=14.3
1. D
2. C
3. A
4. 7.2
5. 14.3
Answer:
the width would be 4 inches and the length would be 6 inches
Answer:
Step-by-step explanation:
Remember all the angles will equal 360 degrees, so add all the other angles together and subtract 360 from the sum of the other three angles for number 5.
5. x = 59 Degrees
6. x = 114 Degrees
