One eighth of the class graduated with an A average, 1/2 got a B, and the rest of the class had a C average. In a class of 800,
2 answers:
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Answer:
(a)
(b) 100
Step-by-step explanation:
Solving (a):
Given
Shape: Rectangle
Required
Calculate the perimeter.
Area is calculated as:
Where
and
Substitute 64.8 for Area
Make L the subject:
Perimeter is calculated as:
Substitute 64.8/W for L
To solve further, we take the derivative of P and set it to 0, afterwards.
Set to 0
Collect Like Terms
Cross Multiply:
Divide through by 2
Take square roots
Recall that:
So:
The perimeter is:
Solving (b):
Given
((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)
Required
Solve
So, the expression becomes:
((1 Group of ) and (1 group of
))/(6 groups of
)
This gives:
((1 Group of ) and (1 group of
))/(6 groups of
)
Group means product, so the expression becomes:
And, as used here means addition
Simplify:
Answer: a) Length of wire to replace old one=9ft
b) Length of wire=3.09ft
Leftover wire=0.91ft
Step-by-step explanation: from the right angle triangle, Tan&=20/15
&=53.13°
Length of wire,x=cos 31.13=x/15
15cos31.13
X=9
b) Tan&=10/6
Tan^-110/6
&=59.05
Length= 6cos59.04
X=3.09
4-3.09=.91= leftover wire
