Pls solve 7th, 8th,9th
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7)) 9x^2-3x-2=0 is the given eqn
dividing the whole by 9 we get
x^2-x/3-2/9
transposing constant to RHS we get
x^2-x/3=2/9
adding 1/36 on both sides then,
x^2-x/3+1/36=2/9+1/36
(x-1/6)^2=2/9+1/36
(x-1/6)^2=9/36
transposing the square it becomes root..
x-1/6=+(or)- 3/6
by solving you will get
2/3,-1/3
done...
9)) we know that tsa of hemisphere is=3pier^2
give diameter is 14 hence radius is half of diameter
therefore radius is 7cm
by substituting you will get
tsa=462cm^s
--------------------------------------------------------
hope it was useful
dividing the whole by 9 we get
x^2-x/3-2/9
transposing constant to RHS we get
x^2-x/3=2/9
adding 1/36 on both sides then,
x^2-x/3+1/36=2/9+1/36
(x-1/6)^2=2/9+1/36
(x-1/6)^2=9/36
transposing the square it becomes root..
x-1/6=+(or)- 3/6
by solving you will get
2/3,-1/3
done...
9)) we know that tsa of hemisphere is=3pier^2
give diameter is 14 hence radius is half of diameter
therefore radius is 7cm
by substituting you will get
tsa=462cm^s
--------------------------------------------------------
hope it was useful
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