please help me in the following questions
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Lipimishra2:
all of them?
Take your time
do except 1,5,8
Kay then!
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I'm doing Q. 2,3,4,6,7,9 as you asked.
Q.2 p(x) = 6x²-x+1 and q(x) = x²+x+1
p(1) = 6(1)²- 1+1 and q(1) = (1)²+1+1
p(1)= 6 and q(1) = 3
A/Q p(1)/q(1) = 6/3 = 2
Q.3. 1/√2 to three decimal places. √2 =1.414
1/1.414 = 1/ (1414/1000) = 1000/1414 = 0.707
Q.4. √243/√3 = √(3×3×3×3×3)/√3 = √(3)⁴ = 3²= 9
Q.6. (28)³+(-15)³+(-13)³
We know, if a+b+c = 0
Then a³+b³+c³- 3abc = 0
= a³+b³+c³=3abc
Checking if a+b+c = 0,
28+(-15)+(-13) = 28 - 28 = 0
So, (28)³+(-15)³+(-13)³ = 3 × 28 × -15 × -13 = 84 × 195 = 16380 (Ans)
Q.7. a+b = 10 and ab = 21, find a³+b³
We've, a+b = 10 and ab= 21
The general formula for a³+b³ = (a+b) (a²+b²-ab). But we don't have the value of a² and b² so we opt for the other formula which is-
a³+b³= (a+b)³- 3ab(a+b)
= 10³- 3× 21 × 10
= 1000 - 630 = 370 (Ans)
Q.9 in attachment
Q.2 p(x) = 6x²-x+1 and q(x) = x²+x+1
p(1) = 6(1)²- 1+1 and q(1) = (1)²+1+1
p(1)= 6 and q(1) = 3
A/Q p(1)/q(1) = 6/3 = 2
Q.3. 1/√2 to three decimal places. √2 =1.414
1/1.414 = 1/ (1414/1000) = 1000/1414 = 0.707
Q.4. √243/√3 = √(3×3×3×3×3)/√3 = √(3)⁴ = 3²= 9
Q.6. (28)³+(-15)³+(-13)³
We know, if a+b+c = 0
Then a³+b³+c³- 3abc = 0
= a³+b³+c³=3abc
Checking if a+b+c = 0,
28+(-15)+(-13) = 28 - 28 = 0
So, (28)³+(-15)³+(-13)³ = 3 × 28 × -15 × -13 = 84 × 195 = 16380 (Ans)
Q.7. a+b = 10 and ab = 21, find a³+b³
We've, a+b = 10 and ab= 21
The general formula for a³+b³ = (a+b) (a²+b²-ab). But we don't have the value of a² and b² so we opt for the other formula which is-
a³+b³= (a+b)³- 3ab(a+b)
= 10³- 3× 21 × 10
= 1000 - 630 = 370 (Ans)
Q.9 in attachment
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you're correct
sorry yyy
it's ok
I think the answer is a=2 and b=-5/root 6
a=-2.....sry
we're not doing it right
a=8......b=root6
No
put it in the equation
Check it now
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