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andrew11 [14]
4 years ago
12

) the length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 3cm/s. when the length is

20 cm and the width is 10 cm, how fast is the area of the rectangle increasing?
Mathematics
1 answer:
VMariaS [17]4 years ago
4 0

It’s a little surprising that this question didn’t come up earlier.  Unfortunately, there’s no intuitive way to understand why “the energy of the rest mass of an object is equal to the rest mass times the speed of light squared” (E=MC2).  A complete derivation/proof includes a fair chunk of math (in the second half of this post), a decent understanding of relativity, and (most important) experimental verification.


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Could someone solve this for me.
Studentka2010 [4]
The answer is 7.6 units
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3 years ago
Kelly spends 3/8 of her free time playing soccer. what percent of her free time was spent playing soccer?
lozanna [386]
I think it's 37.5 . I'm pretty sure cause I think you divide 3/8 . Then you go 2 to the right for your answer with the decimal. Not sure but . I think I'm right.
7 0
3 years ago
The following figure shows A ABC with side lengths to the nearest tenth.
xz_007 [3.2K]

Answer:

17.7

Step-by-step explanation:

angle B = 180 - 81 - 65 = 34 degrees

as the sum of all angles in any triangle is always 180 degrees.

a/sin(A) = b/sin(B) = c/sin(C)

the sides are always on the opposite side of the angle.

so,

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3 0
3 years ago
Jasmine wants to earn $2,000 in simple interest through a
Ierofanga [76]

Answer:

$10000

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First, converting R percent to r a decimal

r = R/100 = 5%/100 = 0.05 per year,

then, solving our equation

P = 2000 / ( 0.05 × 4 ) = 10000

P = $ 10,000.00

The principal required to

accumulate interest of $ 2,000.00

on a rate of 5% per year for 4 years is $ 10,000.00.

3 0
3 years ago
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
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