what is the greatest 5 digit number which when divided by 2, 3, 4, and 5 leaves a remainder of 1 in each case
1 answer:
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First part of question:
Find the general term that represents the situation in terms of k.
The general term for geometric series is:
= the first term of the series
= the geometric ratio
would represent the height at which the ball is first dropped. Therefore:
We also know that the ball has a rebound ratio of 75%, meaning that the ball only bounces 75% of its original height every time it bounces. This appears to be our geometric ratio. Therefore:
Our general term would be:
Second part of question:
If the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces.
represents the initial height:
represents the number of times the ball bounces:
Plugging this back into our general term of the geometric series:
represents the highest height of the ball after 6 bounces.
Third part of question:
If the ball dropped from a height of 235ft, find the total distance traveled by the ball when it strikes the ground for the 12th time.
This would be easier to solve if we have a general term for the <em>sum </em>of a geometric series, which is:
We already know these variables:
Therefore:
Answer:
1. The volume of the cylinder is approximately 0.153 m³
2. 25 cm
3.
4. 16 m
5. 1,785 m³
Step-by-step explanation:
The volume of a solid can be found by the product of the uniform cross-sectional area of the solid and the (continuous) length of the solid
1. The uniform cross-sectional area of the given cylinder = The area of the circle at the base or top
The dimension of the diameter of the circle at the top of the cylinder, d = 50 cm = 0.5 m
The area of the circular cross-section, A = π·d²/4
∴ A = π × 0.5²/4 = 0.0625·π
A = 0.0625·π m²
The height of the cylinder, h = The continuous length of the circular cross-section = 78 cm = 0.78 m
∴ The volume of the cylinder, V = A × h
∴ V = 0.0625·π × 0.78 = 0.04875·π ≈ 0.153
The volume of the cylinder, V ≈ 0.153 m³
2. The given volume of the trapezium, V = 8550 cm³
The length of the short and long parallel sides 'a', and 'b', are 17 cm and 21 cm respectively
The height of the trapezium from the diagram, h = 18 cm
The cross-sectional area of the trapezium, 'A', is found as follows;
A = (17 cm + 21 cm)/2 × 18 cm = 342 cm²
The volume of the trapezium, V = The cross-sectional, A × The (missing) length, 'l' of the trapezium
∴ l = V/A
By substitution, we have;
l = 8550 cm³/(342 cm²) = 25 cm
∴ The Missing Length, l = 25 cm
3. The given volume of the solid having a uniform cross-sectional area is, V = 385 cm³
The area of the (uniform) cross-section of the solid, A = 15 cm²
∴ The length of the solid, 'l', from V = A × l, is given as follows;
l = V/A
∴ l = 385 cm³/(15 cm²) = 25. cm
The length of the solid, l = 25. cm
4. From the diagram, we have;
The cross-sectional area of the solid, A = 216 m²
The length of the solid, l = 16 m
5. The cross-section of the solid can can be assumed to be either;
1. A trapezium from which a rectangle has been removed of dimensions 8 m by 9 m.
2. A triangle located above a rectangle
For scenario one, we have;
The cross-sectional area, A = (12.5 + 9)/2 × 15 - 8 × 9 = 89.25
For scenario two, we find 'A' as follows;
A = 7 × 9 + 1/2 × 15 × 3.5 = 89.25
∴ The cross-sectional area of the solid, A = 89.25 m²
The length, 'l', of the solid, is given as l = 20 m
The volume of the solid, V = A × l
∴ V = 89.25 m² × 20 m = 1,785 m³
The volume of the solid, V = 1,785 m³.