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elena55 [62]
3 years ago
14

A spin balancer rotates the wheel of a car at 500 revolutions per minute. If the diameter of the wheel is 26 inches, what road s

peed is being tested?
a. 26 rad/s
b. 3.2 rad/s
c. 52 rad/s
d. 81 rad/s​
Mathematics
1 answer:
natta225 [31]3 years ago
3 0

Answer:

c. 52 rad/s

Step-by-step explanation:

Final question does not correspond with available option. The real question is: <em>What is the angular speed in radians per second?</em>

At first we assume that spin balance rotates at constant rate and convert given angular speed, measured in revolutions per minute, into radians per second:

\omega = \left(500\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{sec}  \right)

\omega \approx 52.360\,\frac{rad}{s}

Which corresponds to option C.

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astraxan [27]

Answer:

3.14(3.1)^2 = 30.1754

Step-by-step explanation:

pi times radius squared - area of a circle

3 0
3 years ago
Please help ASAP I’ll give brainliest
butalik [34]
Write the equations in matrix,

\left[\begin{array}{ccc}5&-1&1\\1&2&-1\\2&3&-3\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\5\\5\end{array}\right]

Using row transformation,
R₂ <---> R₃ 

\left[\begin{array}{ccc}5&-1&1\\2&3&-3\\1&2&-1\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =  \left[\begin{array}{ccc}4\\5\\5\end{array}\right]

Using,
R₂ ---> R₂ - 2R₃

\left[\begin{array}{ccc}5&-1&1\\0&-1&-1\\1&2&-1\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\-5\\5\end{array}\right]

Using,
R₂ --- > (-1)R₂

\left[\begin{array}{ccc}5&-1&1\\0&1&1\\1&2&-1\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\5\\5\end{array}\right]

Using row transformation,
R₂ <----> R₃
\left[\begin{array}{ccc}5&-1&1\\1&2&-1\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]

Using,
R₂ ---> R₂ - R₁/5

\left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&1&1\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\21/5\\5\end{array}\right]

Using,
R₃ ---> R₃ - 5R₂/11 

\left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&0&17/11\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\21/5\\34/11\end{array}\right]

∴ 5x-y+z = 4 ====(i)
   11y-6z = 21 === (ii)
    17z=34 === (iii)

from iii,
z=2.
Plug z=2 in ii to get y, 
∴y=3.
Plug y and z values in i to get x,
∴x=1

Therefore the solution to the system of equations is (1,3,2)
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There is a body of water that starts with 1 square unit, and doubles in size every day (2 units after 2 days, 4 units after 4 da
Alexxandr [17]

Answer:99 days

Step-by-step explanation:

Given

If one start with one square unit it takes 100 days

i.e. At second day it is 2 unit

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at n th day 2^{n-1}

at 100 day 2^{99} square unit

If we start with 2 square units it would take

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2^n=2^{99}

n=99\ days

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shusha [124]
First box should say “transformed two units left” or something like that
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Klio2033 [76]
$34.65 because of 48.45-13.80
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