Step-by-step explanation:
A)
The length of the box is 30 − 2x inches.
The width of the box is 30 − 2x inches.
The height of the box is x inches.
So the volume is:
V = x (30 − 2x)²
B)
V(3) = 3 (30 − 6)² = 1728
V(4) = 4 (30 − 8)² = 1936
V(5) = 5 (30 − 10)² = 2000
V(6) = 6 (30 − 12)² = 1944
V(7) = 7 (30 − 14)² = 1792
As x increases, the volume of the box increases to a maximum and then decreases.
C)
The ends of the domain occur when V = 0.
0 = x (30 − 2x)²
x = 0 or 15
So the domain is (0, 15).
Here is the solution of the given problem above.
First, let's analyze the question.
Given: 1 bag = 9/10 pound
2/3 bag = ? pound
What we are going to do is to divide 9/10 pound to 3.
so 9/10 divided by 3 and we get 9/30 and to simplify that, 3/10.
So per 1/3 of the bag, there is 3/10 pound.
To get the weight for 2/3 of the bag, we multiply 3/10 by 2 and we get 6/10 or to simplify it, it is 3/5. Therefore, the 2/3 bag weights 3/5 pound. Hope this answer helps.
Use percentage multipliers:
$28/0.8 = $35
The original price of the set is $35.
Hope This Helps!
Answer:
Fraction[Total number of sections filled filled with parents] = 2¹/₁₀ section
Step-by-step explanation:
Given:
Total number of sections filled = 3¹/₂ = 7 / 2 sections
Number of fraction parents watching play = 3/5
Find:
Fraction[Total number of sections filled filled with parents]
Computation:
Fraction[Total number of sections filled filled with parents] = Total number of sections filled x Number of fraction parents watching play
Fraction[Total number of sections filled filled with parents] = [7/2] x [3/5]
Fraction[Total number of sections filled filled with parents] = 21 / 10 sections
Fraction[Total number of sections filled filled with parents] = 2¹/₁₀ section
Answer:
a
Step-by-step explanation:
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