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i think it's a function
i Think that they are put less questions so there is less time to get metals and stuff and this is not a math question.
sketch the graph of y=sin(x+pi/4)
can u help me in this question
is this mathematics?
i searched Bump in wikipedia, it says --->a mathematical function which has a non-empty bounded support
but i don't know really.
ok well can u help me that question : sketch the graph of y=sin(x+pi/4)
i am not able to draw the graph of it Amplitude = 1 , but what is period . can any1 tell me amistre sir ?
its the graph of y=sin(x) shifted pi/4 units to the left
sir why it is shifted left
becasue when we change the value associated with "x" that tends to slide the graph left or right
sir please if u don;t mind can u draw the graph for my help please
draw me the sin(x) basic graph and ill see what i can do
siw why have u denoted 3pi/2 as pi and 7pi/4 as 2 pi approx. ?
i used your graph, and shifted it all to the left by pi/4 and then recalculated the intercepts accordingly to its new values
what was pi is now pi-pi/4 = 3pi/4 what was 2pi is now 2pi-pi/4 = 7pi/4
what was 0, is now 0 - pi/4 = -pi/4
k sir thanks sir if the question was y=sin(x-pi/4) then i would have to shift pi/4 to left instead of right ?
i mean the original orgin would be right to the new origin obtained
the operator sign can be a bit misguided. always go the opposite direction of the sign
x - pi/4 would shift in the+ direction x + pi/4 would shift in the - direction
sir one more problem there was a question in my book : sketch the graph of the following pairs of funcitons on the same axes: 1) y=sinx , y=sin(x+pi/4)
sir what u drew is it the answer of this question , i would have just to draw it by dooted lines to differenciate
so we would sketch (sketch is an approximate drawing as opposed to an exact drawing) the y=sin(x) and then copy it pi/4 to the left yes, dotted would be fine, or a different color I was stuck in post limbo ,,,,
no problem sir , well 10000000000000000s of medals for your helping mind. thanks a lot sir i would keep posting the questions if i would get confused in any topic here . thanks again sir
good luck :)