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4 years ago

What is 63,4721 rounded to 2 decimal places

Mathematics

1 answer:

Eduardwww [97]4 years ago

6 0

It's fairly easy.
634,721.00 would be the answer, if you rounded 634,721 to 2 decimal places.

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A factory produces 2354 hammers every hour. Which expression can be used to find how many hammers the factory produces in 3 hour

ki77a [65]

An expression might be: 2354h / 3hrs. You can make the real division sign if you're using paper. Hope this helped :)

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4 years ago

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The number of views in a cat and a dog after they uploaded are represented by the following

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Answer:

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Step-by-step explanation:

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3 years ago

(b) The area of a rectangular pool is 5488 m\2 If the length of the pool is 98 m, what is its width?

mrs_skeptik [129]

Answer:

Step-by-step explanation:

To find the area of a rectangle we have the foruma A=WxL.

But we already have the area and length so we can plug that in

5488=Wx98

Now its an algebreic expression.

SInce its multiplying we do the opposite, so we divide 98 on both sides.

98/98 crosses itself out so now its 5488/98. Which equals 56. So now our expression is W=56. To fact check we put the numbers 56 and 98 into the formula to see if we get 5488.

A=56x98

A=5488

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2 years ago

The following steps are used to rewrite the polynomial expression x(x-5y) 2y (x-5y

Mice21 [21]

$2x^3y-20x^2y^2+50xy^3$

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3 years ago

Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shap

pantera1 [17]

Answer:

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

Step-by-step explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e $\dfrac{dV}{dt}= 20 \ ft^3/min$

we know that radius r is always twice the diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

then d = h = 2r

h = 2r

r = h/2

The volume of a cone can be given by the formula:

$V = \dfrac{\pi r^2 h}{3}$

$V = \dfrac{\pi (h/2)^2 h}{3}$

$V = \dfrac{1}{12} \pi h^3$

$V = \dfrac{ \pi h^3}{12}$

Taking the differentiation of volume V with respect to time t; we have:

$\dfrac{dV}{dt }= (\dfrac{d}{dh}(\dfrac{\pi h^3}{12})) \times \dfrac{dh}{dt}$

$\dfrac{dV}{dt }= (\dfrac{\pi h^2}{4} ) \times \dfrac{dh}{dt}$

we know that:

$\dfrac{dV}{dt}= 20 \ ft^3/min$

So;we have:

$20= (\dfrac{\pi (15)^2}{4} ) \times \dfrac{dh}{dt}$

$20= 56.25 \pi \times \dfrac{dh}{dt}$

$\mathbf{\dfrac{dh}{dt}= \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

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4 years ago