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[t-nique.com]
Apologies for the confusion. Here it is! Let x = sum of parent's ages y = sum of children's ages n = number of children in the family. x = 2y (Eq1) x-5-5 =4(y-5n) (Eq2) x+15+15 =y+15n (Eq3) (Eq1) vs (Eq2) x = 2y (Eq1) x-5-5 =4(y-5n) (Eq2) . . . y =10n-5 (Eq.i) (Eq1) vs (Eq3) x = 2y (Eq1) x+15+15 =y+15n (Eq3) . . . 2y+30=y+15n . . . y =15n-30 (Eq.ii) (Eq.i) vs (Eq.ii) y =10n-5 (Eq.i) y =15n-30 (Eq.ii) . . . n=5
; [t-nique.com]
Let x = sum of parent's ages y = sum of children's ages n = number of children in the family. x = 2y (1) x-5-5 =4(y-5n) (2) x+15+15 =y+15n (3) (1) vs (2) x = 2y (1) x-5-5 =4(y-5n) (2) . . . y =10n-5 (i) (1) vs (3) x = 2y (1) x+15+15 =y+15n (3) . . . 2y+30=y+15n . . . y =15n-30 (ii) (i) vs (ii) y =10n-5 (i) y =15n-30 (ii) . . . n=5
; [t-nique.com]
Let x = sum of parent's ages y = sum of children's ages n = number of children in the family. x = 2y (1) x-5-5 =4(y-5n) (2) x+15+15 =y+15n (3) (1) vs (2) x = 2y (1) x-5-5 =4(y-5n) (2) . . . y =10n-5 (i) (1) vs (3) x = 2y (1) x+15+15 =y+15n (3) . . . 2y+30=y+15n . . . y =15n-30 (ii) (i) vs (ii) y =10n-5 (i) y =15n-30 (ii) . . . n=5 let me know if you further clarification. dmnatividad@yahoo.com
; [Guest]
bez, it's jacky. let x be the rate of Oscar. y be the rate of james. if they are to meet after 32 seconds of running in opposite direction, that mins that the sum of the distances they've covered is equal to the circumference of the track. try to imagine it.. so, the working equation would be 32x + 32y =440. rate=distance/time. thus, the r of Oscar is 440/60. that is 22/3. so, we now have, 32(22/3) + 32y = 440 we multiply the equation ol thruout by 3 for simplification. so, we now have, 704+96y=1320 y=6.4166.. seconds that is the r of James. but wat is asked hir is the time it will take james to go around the track. so,the formula is t=d/r ryt? t=440/6.4166.. that's around 68. 57 seconds.. got it? :)
; [Guest]
the answer hir shud be letter d. it's 119. 73.. we use the heron's formula. which is the square root of the expression s(s-a)(s-b)(s-b) where s is (a+b+c)/2.
; [Guest]
Its 194 not 192..
; [micxj]
the poster below is not me -micxj
; [micxj]
equatiOn: y=x-3 (1) (y-x)^3=??? (2) solutiOn: substitute equatiOn 2 using equatiOn 1 (y-x)^3= ? ; substitute y=x-3 (x-3-x)^3= ? ; cancel x and -x since it is equal to 0 (-3)^3= -27 means: -3 x -3 x -3 = -27
; [Guest]
in college admission tests, the students are not allowed to use a calculator. so we can't probably know the square root of 190 immediately, but what is asked in the closest approximation so it's not a tough one especially if you know the perfect squares. the given is 190, and the closest perfect square to it is 196. the square root of 196 is 14
; [Guest]
it's ESTIMATION so the answer neeed not be exact. first round off the given numbers then apply the operation 78.35====> 80 38.21=====>40 80(40)=3200 :)
; [mydudz_me]
hey peepz, kindly check your question before your post...
; [mydudz_me]
VERY GOOD....U'RE BRILLIANT...
; [mydudz_me]
simple you have to square root the 190 and the result is 13.78 round of to 14
; [mydudz_me]
please explain further about how to convert decimal to fraction? tnx.
; [mydudz_me]
give me the exact solution of this matter...coz i really don't understand.
; [christinecayanan]
combing? do you play with your hair?
; [Guest]
PAG SURE OI, EVEN ELEM STUDENTS CAN ANSWER THAT QUESTION. HOW COME 8+12=20 IS WRONG??? ARE YOU NUTS????
; [Guest]
i dont understand
; [micxj]
the expression is contained in the ABSOLUTE VALUE sign. so DEFINITELY, the answer is POSITIVE. so the RIGHT answer in 22
; [Guest]
y da.. -22 should be the answer hir..
;
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