Two parallel lines cut by a transversal can create all of the following types of angles EXCEPT
2 answers:
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Usando las relaciones entre velocidad, distancia y tiempo, se encuentra que ella condujo a una velocidad media de 90,5 km/h.
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La <u>velocidad </u><u>es la distancia dividida por el tiempo</u>, por lo que:
- Total de 135,75 km, o sea,
- Llego en 1,5 horas, o sea,
La velocidad es:
División de decimales, o sea, seguimos multiplicando los números por 10 hasta que ninguno sea decimal:
Ella condujo a una velocidad media de 90,5 km/h.
Un problema similar es dado en brainly.com/question/24558377
Answer:
Volume:
Ratio:
Step-by-step explanation:
First of all, we need to find the volume of the hemispherical tank.
The volume of a sphere is given by:
where
r is the radius of the sphere
V is the volume
Here, we have a hemispherical tank: a hemisphere is exactly a sphere cut in a half, so its volume is half that of the sphere:
Now we want to find the ratio between the volume of the hemisphere and its surface area.
The surface area of a sphere is
For a hemisphere, the area of the curved part of the surface is therefore half of this value, so . Moreover, we have to add the surface of the base, which is
. So the total surface area of the hemispherical tank is
Therefore, the ratio betwen the volume and the surface area of the hemisphere is
Answer:
169.04 in² (nearest hundredth)
Step-by-step explanation:
Surface area of a cone = r² +
r
(where r = radius of the base and = slant height)
Given slant height = 10 and surface area = 188.5
Surface area = r² +
r
188.5 = r² + 10
r
r² + 10
r - 188.5 = 0
r = = 4.219621117...
Volume of a cone = (1/3)r²h
(where r = radius of the base and h = height)
We need to find an expression for h in terms of using Pythagoras' Theorem a² + b² = c², where a = radius, b = height and c = slant height
r² + h² = ²
h² = ² - r²
h = √(² - r²)
Therefore, substituting found expression for h:
volume of a cone = (1/3)r²√(
² - r²)
Given slant height = 10 and r = 4.219621117...
volume = 169.0431969... = 169.04 in² (nearest hundredth)