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Q- There are 365 days in the year 1993. The first day is Friday. On what day of the week did 1994 begin?
Anyone?
Anyone?
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Maths, Addmaths and Statistics: Post your doubts here!
- Thread starter OakMoon!
- Start date
365/7 = 52... with remainder "1"
as every 7th day in the figure 7 is the "day b4 friday" i.e. thursday.
thus the remainder "1" is friday which is the 365th day of the year 1993
the next day "saturday" will be the first day of next year 1994.
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Help needed
http://papers.xtremepapers.com/CIE/...Calculator Version) (4024)/4024_w12_qp_12.pdf
Q 18 (whole)
http://papers.xtremepapers.com/CIE/...Calculator Version) (4024)/4024_w12_qp_12.pdf
Q 18 (whole)
i)
taking AQ and AC as base of triangles, tri.AQP and tri.CPQ have the same hieght
so... Area of tri.CPQ = 4h = 6 (where h = hieght)
.................................... h = 6/4 = 1.5
Area of AQP = 2h = 2 x 1.5 = 3 square cm
ii)
tri.CPQ is similar to tri.ABC
linear ratio of the two triangles........... QC : AC
.............................................................. 4 : 6
.............................................................. 2 ; 3
area ratio of the two triangles..... CPQ : ABC
...................................................... 2^2 : 3^2 (as area is the square value, we just need to take the square of linear ratio)
...................................................... 4 : 9
as the ratio is simplified value, so 4y = 6 square cm _(for tri.CPQ)-(where y is an unknown value variable)
y = 6/4 = 3/2 = 1.5
the same way 9y will be the area of tri.ABC
9y = 9 x 1.5 = 13.5
iii) for this part i cant get the answer but if i will i would IA tell you
Hope you understood
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Thanks Yup, I have understood
Yup, I have understood
- Messages
- 858
- Reaction score
- 1,118
- Points
- 153
Thanks Yup, I have understood
Yup, I have understood
- Messages
- 858
- Reaction score
- 1,118
- Points
- 153
Thanks Yup, I have understood
Yup, I have understoodThanks
oww yeah! I got it for part iii)
see.
area of tri.APC = tri.CPQ + tri.AQP
tri.APC = 6 + 3 = 9 sq. cm.
and area of tri.ABC = tri.APC + tri.ABP
................... 13.5 = 9 + tri.ABP
simplify it and you will get "tri.ABP = 4.5"
u r welcome
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Thanks again
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- Messages
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- Reaction score
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- Points
- 523
- Messages
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- Reaction score
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- Points
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Lol.. wait, i forgot to do the steps!
F(x) = 1/2 x(x+1)
F(x-1) = 1/2 (x-1)(x-1+1) = 1/2 (x-1)(x)
F(x) - F(x-1) = 1/2 x(x+1) - 1/2 (x^2 -x) [x^2 means x squared]
= 1/2 x^2+x - 1/2 (x^2 -x)
= 1/2 [(x^2 + x) - (x^2 -x)] ......... Took 1/2 common.
= 1/2 (x+x)
= 1/2 2x
F(x) = 1/2 x(x+1)
F(x-1) = 1/2 (x-1)(x-1+1) = 1/2 (x-1)(x)
F(x) - F(x-1) = 1/2 x(x+1) - 1/2 (x^2 -x) [x^2 means x squared]
= 1/2 x^2+x - 1/2 (x^2 -x)
= 1/2 [(x^2 + x) - (x^2 -x)] ......... Took 1/2 common.
= 1/2 (x+x)
= 1/2 2x
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I just have 4 Oxford Maths books for O'levels and MyMaths; good enough for Additional Maths Olevels?
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I just have 4 Oxford Maths books for O'levels and MyMaths; good enough for Additional Maths Olevels?
no! for add maths u better take classes from some teacher. but the book i used in school was "New Additional Mathematics" by Ho Soo Thong. but now my new teacher uses the "classified" to teach us.
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MyMaths teaches till Alevel maths...so dont you think it is good enough?
es he/she can be good.