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TEA [102]
3 years ago
13

So i need help with the question below.

Mathematics
1 answer:
Shkiper50 [21]3 years ago
8 0

Answer:

Option (1)

Step-by-step explanation:

Option (1)

3\frac{3}{9}+7\frac{6}{11}=10\frac{87}{99}

3\frac{3}{9}+7\frac{6}{11} =3+\frac{3}{9}+7+\frac{6}{11}

             =(3+7)+(\frac{3}{9}+\frac{6}{11})

             =10+(\frac{3}{9}\times \frac{11}{11})+(\frac{6}{11}\times \frac{9}{9})

             =10+(\frac{33}{99}+\frac{54}{99})

             =10+\frac{87}{99}

             =10\frac{87}{99}

True.

Option (2)

2\frac{3}{8}+6\frac{4}{5}=8\frac{12}{40}

2\frac{3}{8}+6\frac{4}{5}=2+\frac{3}{8}+6+\frac{4}{5}

             =(2+6)+(\frac{3}{8}+\frac{4}{5})

             =(8)+(\frac{3}{8}\times \frac{5}{5} +\frac{4}{5}\times \frac{8}{8})

             =8+(\frac{15}{40}+\frac{32}{40})

             =8+\frac{47}{40}

             =8+\frac{40}{40}+\frac{7}{40}

             =8+1+\frac{7}{40}

             =9+\frac{7}{40}

             =9\frac{7}{40}

False.

Option(3)

3\frac{3}{7}+4\frac{2}{3}=7\frac{2}{21}

3\frac{3}{7}+4\frac{2}{3}=(3+\frac{3}{7})+(4+\frac{2}{3})

             =(3+4)+(\frac{3}{7}+\frac{2}{3})

             =7+(\frac{3}{7}\times \frac{3}{3} +\frac{2}{3}\times \frac{7}{7})

             =7+(\frac{9}{21}+\frac{14}{21})

             =7+(\frac{23}{21})

             =7+(\frac{21}{21}+\frac{2}{21})

             =8+\frac{1}{21}

             =8\frac{1}{21}

False.

Option (4)

4\frac{5}{6}+5\frac{5}{7}=9\frac{10}{13}

4\frac{5}{6}+5\frac{5}{7}=4+\frac{5}{6}+5+\frac{5}{7}

             =(4+5)+(\frac{5}{6}+\frac{5}{7})

             =9+(\frac{5}{6}\times \frac{7}{7}+\frac{5}{7}\times \frac{6}{6})

             =9+(\frac{35}{42}+\frac{30}{42})

             =9+\frac{65}{42}

             =9+(\frac{42}{42}+\frac{23}{42})

             =9+1+\frac{23}{42}

             =10\frac{23}{42}

False.

Therefore, Option (1) is the answer.

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A car driving at 20m/s accelerates at 3m/s2 for 3s. What is its final velocity? How far will it travel during this time?
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Answers:

Final velocity = 29 m/s

Distance traveled = 73.5 meters

========================================================

Work Shown:

The given data is

v_i = 20 \text{ m}/\text{s} = \text{initial velocity}\\\\a = 3 \text{ m}/\text{s}^2 = \text{acceleration}\\\\t = 3 \text{ s} = \text{time duration}\\\\

This leads to

v_f = v_i + a*t\\\\v_f = 20 + 3*3\\\\v_f = 20 + 9\\\\v_f = 29\\\\

The final velocity is 29 m/s.

----------

We can use that final velocity to find the distance traveled.

x = 0.5*(v_i+v_f)*t\\\\x = 0.5*(20+29)*3\\\\x = 0.5*(49)*3\\\\x = 24.5*3\\\\x = 73.5\\\\

The distance traveled is 73.5 meters.

----------

An alternative way to calculate the distance is to do this

x = \frac{(v_f)^2 - (v_i)^2}{2a}\\\\x = \frac{(29)^2 - (20)^2}{2*3}\\\\x = \frac{841 - 400}{6}\\\\x = \frac{441}{6}\\\\x = 73.5\\\\

Or you could do this

x = v_i*t + 0.5*a*t^2\\\\x = 20*3 + 0.5*3*3^2\\\\x = 20*3 + 0.5*3*9\\\\x = 60 + 13.5\\\\x = 73.5\\\\

For more information, check out the Kinematics equations.

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