Question

3.a Distance between Shameer's house and Chayan's house is 405km. If Chayan leaves at
10.00am for Shameer's house by bike and reaches, there at 7.00pm. Then find out his
speed.
b. Solve the following:
78 - (-34) + 12-52 + (-19)
Solve 23/56 + 3/8 - 15/28​

Answer 1

Answer:

the answer is very simple Sameer house and charging houses 4051 bilever 10 year eve in their binary j7 the speed of the bike 2019

Answer 2

a.

t1 =7:00 pm = 19:00pm

t2 =10:00 am = 10:00 am

time =t1 - t2

= (19 - 10) hrs

= 9 hrs

distance = 405 km

$speed \: = \frac{distance}{time} \\\:\:\: = \frac{405}{9} \\\:\:\: = 45 \: km \: {hr}^{ - 1}$

therefore,

speed is 45 km/hr

b.

78 - (-34) + 12-52 + (-19)

$= 78 - ( - 34) + 12 - 52 + ( - 19) \\ = 78 + 34 + 12 - 52 - 19 \\ = 53$

Solve 23/56 + 3/8 - 15/28

$= \frac{23}{56} + \frac{3}{8} - \frac{15}{28} \\ \\ lcm \: of \: 56 \: 8 \: and \: 28 \: is \: 168 \ \\ \ \\ = \frac{23 \times 3}{56 \times 3} + \frac{3 \times 21}{8 \times 21} - \frac{15 \times 6}{28 \times 6} \\ \\ = \frac{69}{168} + \frac{63}{168} - \frac{90}{168} \\ \\ = \frac{69 + 63 - 90}{168} \\ \\ = \frac{132}{168} \\ \\ = \frac{33}{42}$

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