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Maths, Addmaths and Statistics: Post your doubts here!

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Angle B is 53.4 correct to the nearest tenth of a degree (0.1)

0.1/2 = 0.05
53.4 + 0.05 = 53.45 --> UB
53.4 - 0.05 = 53.35 --> LB

Upper Bound of Angle C so Angles A and B must be smallest (LB) to get highest C.
Upper bound of C = 180 - LB of A - LB of B
UBC = 180 - 53.35 - 61.5 => 65.15
Nearest tenth of a degree? Isn't it supposed to be 10? :/
 
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distance travelled in the first 20 sec
1/2 (20)(u) =10u (area under the graph)
D=10u.....> eq 1
the car has travelled 2D meters at time t seconds from the start.
i.e distance travelled in the first t seconds =area under trapezium
2D= 1/2 (u)(t+t-20)
substitute the value of D from eq 1 and solve ull get the answer
 
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Q24:
10268579_790516840972308_5704294759694203166_n.jpg

At first i thought to make a bisector of A,but before uploading i reread it....you can see for yourself.
 
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ms.JPG venn.JPG
heres a question with its m.s i dont know where m i making mistake but in the last part of the question i dont think so 4 would be included but it is here in the marking scheme
can anybody solve the last part only ??o_O plz
 
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