5 fractions that are near but not equivalent to 2/3

1 answer:

Montano1993 [528]3 years ago

5 0

Five fractions that are near to the fraction two-thirds but not equivalent or 1/2 1/4 3/6 5/9 15/21.

Hope this helps!!!!

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A ream of paper contains 500 sheets. how many are in 31 reams

sweet [91]

There are 15,500 sheets of paper

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3 years ago

Find the flow rate through a tube of radius 6 cm, assuming that the velocity of fluid particles at a distance r cm from the cent

vaieri [72.5K]

Answer:

791.68 cm/s

Step-by-step explanation:

The volume flow rate can be interpreted as the integral of fluid velocity over area

$\dot{V} = \int\limits^6_0 {v(r) 2\pi r} \, dr\\\dot{V} = 2\pi\int\limits^6_0 {(25-r^2)r} \, dr\\\dot{V} = 2\pi\int\limits^6_0 {25r-r^3} \, dr\\\\\dot{V} = 2\pi[12.5r^2 - r^4/4]_0^6\\\dot{V} = 2\pi(12.5*6^2 - 6^4/4 - 12.5*0 - 0)\\\dot{V} = 2\pi*126 = 791.68 cm/s$

7 0

3 years ago

Convert 5.764764764 … to a rational expression in the form of a over b, where b ≠ 0

zubka84 [21]

I just learned a short time ago how to convert a repeating decimal to a fraction.

-- Take the repeating part of the decimal. In this one, it's '764' .

-- Make a fraction out of it by writing it over the same number of 9s.

   The fraction here is 764/999 .

-- Simplify it if possible and if you feel like it.

   764/999 can't be simplified. (I think.)

So the rational expressions for this decimal are

   (5 and 764/999)   or   5759/999   .

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3 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
\\\\\\
\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
\\\\\\
-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
\\\\\\
\textit{the initial \underline{position} was 400ft away at 0secs}
\\\\\\
-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
\\\\\\
25=t^2\implies \boxed{5=t}$

part B) check the picture below

5 0

3 years ago

Some one plz help me

Deffense [45]

Answer:

Given:

- angle: E

- opposite-leg : 3

- adjacent - leg: 4

- hypotenuse: 5

sin(E) = opposite-leg / hypotenuse = 3 / 5

Answer: 3 / 5

Step-by-step explanation:

Brainliest?

5 0

3 years ago

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