For 5 you would need 1,000 ml of water and 7 you need 1,400 ml
Answer:
5.1 cm² approximately
Step-by-step explanation:
A lighter= Area inscribed -1/2r² sin80°
A lighter=π r² (80°/360°) -1/2(5 cm)² sin 80°
A lighter=π(5cm)² (80°/360°) -1/2(5 cm)² sin 80°
A lighter=3.14*25cm²(80°/360°) - 12.5cm² sin 80°
A lighter=17.444…cm² - 12.31009... cm²
A lighter=5.13cm² approximately
Answer:
Step-by-step explanation:
Let's solve your equation step-by-step.
36p=5p(7.5)
Step 1: Simplify both sides of the equation.
36p=37.5p
Step 2: Subtract 37.5p from both sides.
36p−37.5p=37.5p−37.5p
−1.5p=0
Step 3: Divide both sides by -1.5.
−1.5p
−1.5
=
0
−1.5
p=0
How fast the volume of the sphere is changing when the surface area is 10 square centimeters is it is increasing at a rate of 30 cm³/s.
To solve the question, we need to know the volume of a sphere
<h3>
Volume of a sphere</h3>
The volume of a sphere V = 4πr³/3 where r = radius of sphere.
<h3>How fast the volume of the sphere is changing</h3>
To find the how fast the volume of the sphere is changing, we find rate of change of volume of the sphere. Thus, we differentiate its volume with respect to time.
So, dV/dt = d(4πr³/3)/dt
= d(4πr³/3)/dr × dr/dt
= 4πr²dr/dt where
- dr/dt = rate of change of radius of sphere and
- 4πr² = surface area of sphere
Given that
- dr/dt = + 3 cm/s (positive since it is increasing) and
- 4πr² = surface area of sphere = 10 cm²,
Substituting the values of the variables into the equation, we have
dV/dt = 4πr²dr/dt
dV/dt = 10 cm² × 3 cm/s
dV/dt = 30 cm³/s
So, how fast the volume of the sphere is changing when the surface area is 10 square centimeters is it is increasing at a rate of 30 cm³/s.
Learn more about how fast volume of sphere is changing here:
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