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pishuonlain [190]
3 years ago
6

How to do this Question a please help from the smallest to the biggest

Mathematics
2 answers:
IrinaK [193]3 years ago
7 0
To solve this type of problems, the easiest way to compare is to change them all to fractions, or to decimals, whichever is easier for you.

For the first one, change them all to decimals,
A. 525/1000=0.525
B. 3/4 = 0.750
C. 0.55 =0.550
Thus it will be obvious, whether you'd like smallest to biggest or otherwise.

For the second problem, 
A. 3.85=3.850
B. 3 5/8 = 3.625
C. 3.805 = 3.850

Choose the smallest to biggest.



Pachacha [2.7K]3 years ago
5 0
A. turn all fractions into decimals or all decimals into fractions; I've decided to turn all fractions into decimals

525
------- = 0.525
1000

3
-- = 0.75 = 0.750 ~ (add a zero at the end to make the decimal places         4                                equal with 0.525)

0.55 = 0.55 = 0.550 ~<span>(add a zero at the end to make the decimal places                                         equal with 0.525 and 0.750)
</span>
<span>                                          Answer to Problem 'a':</span>

Smallest = 0.525 = 525
                              -------
                              1000

Middle = 0.550 = 0.55

Largest = 0.750 = 3
                              --
                              4

b. <span>turn all fractions/mixed fractions into decimals or all decimals into fractions; I've decided to turn all fractions/mixed fractions into decimals
</span>
3.805 = 3.805

3.85 = 3.850 ~ ( <span>add a zero at the end to make the decimal places equal                                with 3.805)
</span>
3 4/5 = 80
           ----- = 0.80  =0.800 ~ (add a zero at the end to make the decimal
           100                              places equal with 3.805 and 3.850)

                                 Answers to Problem 'b':

Smallest = 0.800 = 0.80 = 3 4/5

Middle = 3.805 = 3.805

Largest = 3.850 = 3.85

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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height
Crank

Answer:

a) h = 0.1: \bar v = -11\,\frac{ft}{s}, h = 0.01: \bar v = -10.1\,\frac{ft}{s}, h = 0.001: \bar v = -10\,\frac{ft}{s}, b) The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

Step-by-step explanation:

a) We know that y = 30\cdot t -10\cdot t^{2} describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity (\bar v), measured in feet per second, can be done by means of the following definition:

\bar v = \frac{y(2+h)-y(2)}{h}

Where:

y(2) - Position of the ball evaluated at t = 2\,s, measured in feet.

y(2+h) - Position of the ball evaluated at t =(2+h)\,s, measured in feet.

h - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

h = 0.1\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}

y(2.1) = 18.9\,ft

\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}

\bar v = -11\,\frac{ft}{s}

h = 0.01\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}

y(2.01) = 19.899\,ft

\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}

\bar v = -10.1\,\frac{ft}{s}

h = 0.001\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}

y(2.001) = 19.99\,ft

\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}

\bar v = -10\,\frac{ft}{s}

b) The instantaneous velocity when t = 2\,s can be obtained by using the following limit:

v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}

v(t) =  \lim_{h \to 0} 30-20\cdot t-10\cdot h

v(t) = 30\cdot  \lim_{h \to 0} 1 - 20\cdot t \cdot  \lim_{h \to 0} 1 - 10\cdot  \lim_{h \to 0} h

v(t) = 30-20\cdot t

And we finally evaluate the instantaneous velocity at t = 2\,s:

v(2) = 30-20\cdot (2)

v(2) = -10\,\frac{ft}{s}

The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

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Tresset [83]

\huge \green{ \boxed{ANSWER}}

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USE CALCULATOR -_-

\mathfrak{JazmineChoi}

<em><u>STAN </u></em><em><u>TREASURE</u></em>

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