Divide 575 by 14 by using partial quotients
1 answer:
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The average density of the rod is 0.704 kg/m.
For given question,
We have been given the linear density in a rod 5 m long is 10 / x + 4 kg/m, where x is measured in meters from one end of the rod.
We need to find the
The length of rod is, L = 5 m.
The linear density of rod is, ρ = 10/( x + 4) kg/m
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 5,
The expression for the average density is given as,
⇒ ρ'
......................(1)
Using u = x + 4
du = dx
u₁ = x₁ + 4
u₁ = 0 + 4
u₁ = 4
and
u₂ = x₂ + 4
u₂ = 5 + 4
u₂ = 9
By entering the values above into (1), we have:
⇒ ρ'
= 0.704
Thus, we can conclude that the average density of the rod is 0.704 kg/m.
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The ratio of the area of the smaller rectangle to the area of the larger
rectangle is
Step-by-step explanation:
The formula of the area of a rectangle is A = l × w, where l and w
are its dimensions
- The dimensions of a smaller rectangle are 3 ft by 9 ft
- The dimensions of a larger rectangle are 5 ft by 15 ft
We want to find the ratio of the area of the smaller rectangle to
the area of the larger rectangle
∵ The dimensions of the smaller rectangle are 3 ft and 9 ft
∴ The area of the smaller rectangle = 3 × 9 = 27 feet²
∵ The dimensions of the larger rectangle are 5 ft and 15 ft
∴ The area of the smaller rectangle = 5 × 15 = 75 feet²
Let us find the ratio of the area of the smaller rectangle to the area
of the larger rectangle
→ smaller rectangle : larger rectangle
→ 27 : 75
Divide the both terms of the ratio by 3
→ 9 : 25
The two terms of the ratio do not divisible by any other number
∴ The simplest form of the ratio is 9 : 25
∴ The ratio of the area of the smaller rectangle to the area of the larger
rectangle =
The ratio of the area of the smaller rectangle to the area of the larger
rectangle is
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You can learn more about the ratio in brainly.com/question/10781917
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