Simplify the following expression.
2 answers:
You might be interested in
Answer
f(t)=2cos(t)
Step-by-step explanation:
let's describe the situation first, a fly is starting 2 meters from the bulb and flies towards the bulb and when it is really close to the bulb, it flies 2 meters further away from the bulb this means that it reaches d = 0m (distance from the bulb) and then flies further 2 meters so d = -2m.
it this point the fly returns back and touches the bulb and flies away (ends it's oscillatory motion ). d = 0 again and story ends here.
here if we want to model this problem with time function , the cosine function seems the best fit with amplitude of 2, so the answer is f(t) = 2cos(t).
Now you can ask why cosine function? well if you look at the graph or the plot of the function it perfectly captures the physical situation going on here in this problem.
Domain is 0 to 2 because it is one complete cycle and the range is -2 to +2.
Answer:
Step-by-step explanation:
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.