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Sure,Could you please explain the Range bit with some more detail please?
"− 2.5 ≤ x < 0"
Put x as 1 in f(x) you get -2.5 as lower range
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Sure,Could you please explain the Range bit with some more detail please?
"− 2.5 ≤ x < 0"
Sure,
Put x as 1 in f(x) you get -2.5 as lower range
how did you get y = -x/2 + pi .... we are supposed to get y = (pi-x)/2On the same graph, draw the line y = -x/2 + pi
When x = 0 y = pie, when x = 3 y = 1.64 now connect both the points
ZaqZainabRight. I get that, but why do we put it in this format? -2.5<= x <0
Where did the 0 come from?
Post the question's link.how did you get y = -x/2 + pi .... we are supposed to get y = (pi-x)/2
which question?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_12.pdfwhich question?
Try substituting values for more than 1 and you'll find that the values of range keep decreasing till zero and it will never touch it and by knowing the graph of the function you will be able to relate what I'm sayinghttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf
question 9iii) How is the domain of f inverse has o ?
My mistake.
but we will get answers like 1.bla bla bla 0. bla bla bla .. how are we supposed to plot this on the graph there must be mistake somewhere plus the graph is not given in the ms so IDK WHAT TO DOMy mistake.
So, put x = 0 and get y from the eqn, and put any other value of x and get your y, and plot the graph
Take very far points on x axis and connect 'embut we will get answers like 1.bla bla bla 0. bla bla bla .. how are we supposed to plot this on the graph there must be mistake somewhere plus the graph is not given in the ms so IDK WHAT TO DO
omg are you serious ? .. since when do we get these kind of decimals .. ABNORMAL decimals !!Take very far points on x axis and connect 'em
Its the only way I do it, if you get more accurate way, do tell meomg are you serious ? .. since when do we get these kind of decimals .. ABNORMAL decimals !!
What? this is the formula
Phew, that took a while to write clearly and illustrating properly sorry for the delay
View attachment 41014 View attachment 41015 View attachment 41016
Ikr.What? this is the formula
How did you get 3,9 ?I dont know this method is correct or not
We had to show a point of reflection, hence we have x1,y1 as (-1,3) and we found a midpoint (3,9) by sim. equation,
So, as to find midpoint we do (x1+x2)/2, and (y1+y1)/2,
similarly,
We know have to find x2 and y2, we have x1 y1 and midpoint
so (x1,y1, + x2,y2)/2 = midpoints
=(-1,3 + x2,y2) = 2(3,9)
=(-1,3 + x2,y2) = (6,18)
x2=(6-(-1))=7
y2=(18-3)=15
so reflection on (7,15)
T H I S R E A L L Y W O R K S
Yes I nees work solution on question 8
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