Add Math Qtns
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Depends on question.Originally posted by Y_Shun:
Where is the sign?Originally posted by Y_Shun: -
Your question not very clear.
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It says...Originally posted by eagle:Depends on question.
Write down three matries whose sum shows the total number of cusomters in all three restaurant branches who purchased each type of meal on the particular day. Evaluate this total. -
Originally posted by Y_Shun:
wrong. A row vector is not a column vector. Try inputting a row vector as a column vector in a computer and your result will be wrong.
[b]
Stricly speaking, its wrong(the equation you put up there) but some teachers close one eye. -
Q1. not equal
Q2. plus minus multiply or divide???? -
wait ar...the thing sot liaoOriginally posted by eagle:Where is the sign?
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good answerOriginally posted by HyuugaNeji:wrong. A row vector is not a column vector. Try inputting a row vector as a column vector in a computer and your result will be wrong.
Stricly speaking, its wrong(the equation you put up there) but some teachers close one eye. -
both signs are negative.
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it still depends on where you want to put to restuarants (left hand side)(top)Originally posted by Y_Shun:It says...
Write down three matries whose sum shows the total number of cusomters in all three restaurant branches who purchased each type of meal on the particular day. Evaluate this total.
and types of meals. (top)(left hand side) -
so its correct or wrong??? it depends on me or what....?Originally posted by HyuugaNeji:it still depends on where you want to put to restuarants (left hand side)(top)
and types of meals. (top)(left hand side) -
At A maths level, you will be marked right. No guarantee though.Originally posted by Y_Shun:so its correct or wrong??? it depends on me or what....?
It depends on how you define the left and top side. If you define restuarants on the left will become figure A vector.
But if you choose to define type of meals on the left, your restuarant has to be defined on the top, then it will be a figure b vector.
Anyway, most pple just follow what is defined in the question.
For question 2, you can square both sides since these are real numbers.
x^4 - 2x^2 + 1 = x^2 - 2x + 1
x^4 - 3x^2 +2x=0
x(x^3-3x+2)=0
x=0 x=-2 x=1 -
Is that symbol for absolute value?Originally posted by Y_Shun: -
yaOriginally posted by lpx88:Is that symbol for absolute value? -
I think TS did thisOrxiginally posted by HyuugaNeji:For question 2, you can square both sides since these are real numbers.
x^4 - 2x^2 + 1 = x^2 - 2x + 1
x^4 - 3x^2 +2x=0
x(x^3-3x+2)=0
x=0 x=-2 x=1
(x²-1)² = (x-1)²
(x+1)²(x-1)² - (x-1)² = 0
(x-1)²(x)² = 0
which is wrong
underlined one should then be
(x-1)²(x² + 2x + 1 - 1) = 0
(x-1)²(x)(x+2) = 0
and you get your answer -
absolute value i thought cant have negative values one?Or im wrong?Originally posted by Y_Shun:ya
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seriously, i never even square it leh..i juz remove the aboslute and solve it >_>
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inside the absolute can have negative. After doing the absolute, cannot have negative.Originally posted by lpx88:absolute value i thought cant have negative values one?Or im wrong?
In this eg, if x=-2
answer will be |3|=|-3| -
then you lose a solutionOriginally posted by Y_Shun:seriously, i never even square it leh..i juz remove the aboslute and solve it >_>
cuz if you want to do that
x²-1 = x-1 or x²-1 = -(x-1)
for x²-1 = x-1
x = 0 or x = 1
for x²-1 = -(x-1)
x=1 or x = -2
you lose the x=-2 solution -
abs(x²-1)=abs(x-1)
two absolute functions can be translated as:
f(x)=g(x) or -f(x)=-g(x), and
f(x)=-g(x) or -f(x)=g(x)
x²-1=x-1
x²-x=0
x(x-1)=0
x = 0 or 1
1-x²=x-1
x²+x-2=0
(x+2)(x-1)=0
x = 1 or -2