Let X represent the number of hours you travelled
Let B represent number of hours Brian travelled
Equation 1 is X + B = 20 (since combination of number of hours you and Brian travelled equal to 20 hours)
Equation 2 is M = 5B (since you travelled five times more than the amount in which Brian travelled)
Plug Equation 2 into Equation 1 (replacing M with its value)
5B + B = 20
Rearrange equation to make B its subject (find B)
6B = 20
6B/6 = 20/6
B = 3(1/3)
Brian travelled a total of 3(1/3) (three and a third) hours.
(1/3) of an hour = 60/3 or 60*(1/3) = 20mins
Brian travelled a total of 3hrs & 20mins.
Plug B value back into Equation 1
M + B = 20
M + 3(1/3) = 20
Rearrange equation to make M its subject (find M)
M = 20 - 3(1/3) = 16(2/3)
You travelled a total of 16(2/3) (sixteen and two-thirds) hours.
(2/3) of an hour = 60/3*2 or 60*(2/3) = 40mins
You travelled a total of 16hrs & 40mins.
Answer:
Yes since 40+30=70 so it would be the same
Step-by-step explanation:
<h3>
<u>Answer</u><u>:</u></h3>
<u>➲</u><u> </u><u>(</u><u> </u><u>1</u><u> </u><u>+</u><u> </u><u>sec </u><u>x </u><u>)</u><u>(</u><u> </u><u>cosec </u><u>x </u><u>-</u><u> </u><u>cot </u><u>x </u><u>)</u><u> </u><u>=</u><u> </u><u>tan </u><u>x</u>
- <em>Solving</em><em> </em><em>for </em><em>L.H.S</em>








<u>➲</u><u> </u><u>(</u><u> </u><u>1</u><u>+</u><u> </u><u>sec </u><u>x </u><u>)</u><u>(</u><u> </u><u>1</u><u>-</u><u> </u><u>cos </u><u>x </u><u>)</u><u> </u><u>=</u><u> </u><u>sin </u><u>x.</u><u> </u><u>tan </u><u>x</u>
- <em>Solving </em><em>for </em><em>L.H.S</em>







Heyyyy
how r u
sorry for being annyoing but i won’t let me ask a question so
The value of

, and

of each data is shown in the diagram below
Group A
Mean =

Median = 49.5
Range =

Interquartile Range =

Group B
Mean =

Median = 50
Range = 62-40=22
Interquartile range = 58.5-42 = 16.5
Correct statement: the time in Group B has a greater interquartile range