3 years ago

The volume of a cube is 1, 953.125 centimeters cubed. Find the length of the sides. Round to the nearest tenth.

Mathematics

1 answer:

Shtirlitz [24]3 years ago

6 0

1953.13 is the length of the sided

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The lunch special at Jimmy John’s costs $5.60. The math club has $50.40 in its treasury. How many lunch specials can the club bu

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Step-by-step explanation:

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3 years ago

A video rental store has 6,000 movies. One Friday, 3/5 of the movies were rented. How many movies were rented that Friday night?

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3 years ago

Read 2 more answers

An ice cube is melting, and the lengths of its sides are decreasing at a rate of 0.8 millimeters per minute At what rate is the

julia-pushkina [17]

Answer:

The rate of decrease is: $43.2mm^3/min$

Step-by-step explanation:

Given

$l = 18mm$

$\frac{dl}{dt} = -0.8mm/min$ ---- We used minus because the rate is decreasing

Required

Rate of decrease when: $l = 18mm$

The volume of the cube is:

$V = l^3$

Differentiate

$\frac{dV}{dl} = 3l^2$

Make dV the subject

$dV = 3l^2 \cdot dl$

Divide both sides by dt

$\frac{dV}{dt} = 3l^2 \cdot \frac{dl}{dt}$

Given that: $l = 18mm$ and $\frac{dl}{dt} = -0.8mm/min$

$\frac{dV}{dt} = 3 * (18mm)^2 * (-0.8mm/min)$

$\frac{dV}{dt} = 3 * 18 *-0.8mm^3/min$

$\frac{dV}{dt} = -43.2mm^3/min$

<em>Hence, the rate of decrease is: 43.2mm^3/min</em>

8 0

3 years ago

Help me, Thanks!!

Marrrta [24]

2.) dT/dt = -t^2 + 3t

dT = (-t^2 + 3t)dt

T = -t^3/3 + 3t^2/2 + C

at 4:00 AM

80 = -4^3/3 + 3*4^2/2 + C

C = 80 + 4^3/3 - 3*4^2/2

C = 77.33

T = -t^3/3 + 3t^2/2 + 77.33

at 8:00 AM

T = -8^3/3 + 3*8^2/2 + 77.33

T = 2.67 deg

5 0

3 years ago