Richard can read 1/4 of a book in 3/5 of an hour. At this rate, how much can Richard read in one hour

1 answer:

Volgvan3 years ago

3 0

3/5 hour ⇒ 1/4 of the book

Find 1/5 hour:
1/5 hour ⇒ 1/4 ÷ 3 = 1/4 x 1/3 = 1/12 of the book

Find 5/5 (1 hour):
5/5 hour ⇒ 1/12 x 5 = 5/12 of the book

Answer: 5/12 of the book

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A right circular cylinder is inscribed in a sphere of radius r. Find the dimensions of such a cylinder with the largest possible

Burka [1]

Answer:

Largest volume in terms of r = 4r^3π/ 3√3 or you could write it as 4√3r^3π/9.

Step-by-step explanation:

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3 years ago

Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an

Natasha_Volkova [10]

Answer:

The volume of the solid is 243 $\sqrt{2} \ \pi$

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0 ≤ ρ² ≤ 81

0 ≤ ρ ≤ 9

The intersection that takes place in the sphere and the cone is:

$x^2 +y^2 ( \sqrt{x^2 +y^2 })^2 = 81$

$2(x^2 + y^2) =81$

$x^2 +y^2 = \dfrac{81}{2}$

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

$z = \sqrt{x^2+y^2}$

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; $\dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}$

Thus, volume: $V = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4} \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho ^2 \ sin \phi \ d\rho \ d \theta \ d \phi$