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How to do this question?
Find the coordinates of the point where the line 2y = 3x + 15 crosses the y-axis.
Find the coordinates of the point where the line 2y = 3x + 15 crosses the y-axis.
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How to do this question?
Find the coordinates of the point where the line 2y = 3x + 15 crosses the y-axis.
suppose the image of ABC by rotation is A'B'C' then join all the corresponding points as A to A' ,B to B'.PLEASE SOMEONE HELP ME IN CENTER OF ROTATION. PLEASE.
for this type of questions the method is samethankuuuu so much can u help me with this one too
meiling can complete a piece of work in 12 days and suling can complete it in 18 days .findthe number of days in which both,working together will take to complete the work
a)AR= AB+BRCan somebody help solving this! the last two parts!View attachment 27569
not in trapeziumare the opposite angle in EVERY quadrilateral supplementary? and is there a set of notes available which addresses the 'RATTA' topics lyk angle properties, lines of symmetry order of rotation symmetry etc?
When Line passes through Y-Axis then x=0
So, 2y=3(0)+15
2y=15
Y=7.5.
Therefore, co-ordinates = (0,7.5)
thanks, but how do you know that when line passes through Y-axis, then x=0??
not in trapezium
thanks, but how do you know that when line passes through Y-axis, then x=0??
suppose the image of ABC by rotation is A'B'C' then join all the corresponding points as A to A' ,B to B'.
then find the bisector of each line. the point where the bisectors intersect is centre of rotation.
2ND LAST ONE IS P+K/2 Qa)AR= AB+BR
=p+k(q/2)
post the answers of 1st two so i can answer the last part as i m too exhausted to attempt the other parts as well
THEN IT WILL BE CORRECT BUT NOT OUTSIDEwhat if a trapezium is made inside a circle??(cyclic quadrilateral)
by the way what issue do you have with my caps lockThanks. Turn off your caps lock. It's annoying.
for this type of questions the method is same
1/12+1/18=1/x
x= no of days together
an easier way would be to just make pairs of arcs from corresponding vertices of the triangle, and make the perpendicular bisectors that way. There is lesser chance of error, and less time wasted
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