Quick equation help Watch

tennantgirl
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#1
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Hello all
I have a quick question on a homework sheet which I am a little stuck on how to work it out- trust me I have tried!
so it's...

1) you are given the identity x² -12x+10= (x=a)² +b
work out the values of a and b

2) The diameter of the base of the cone is xcm. The height of the cone is also xcm. The volume of the cone is Vcm³.
Find the formula for x in terms of V and x.

So far I have got V=1/3 ? 1/2x²x ...but I don't know what to do now.

Any help would be really appreciated, as I've been stuck on these questions for ages. Thank you!
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JLKeating
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(Original post by tennantgirl)
Hello all
I have a quick question on a homework sheet which I am a little stuck on how to work it out- trust me I have tried!
so it's...

1) you are given the identity x² -12x+10= (x=a)² +b
work out the values of a and b
I got a=6 because you multiply out (x-a)(x-a) and so you get the identity x²-12x+10 = x²-2ax-a²+b and then we were taught to write out how many x²'s, x¹'s and xº's so for the left side you would have 1, -2a and -a²+b and then the right hand side you would get 1, -12 and 10. So -2a = -12, -a = -6 therefore a must be 6. Then b would be -26 if you put it into the equation of -a²+b = 10. I think
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Nalced
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I feel so stupid trying to understand this thread. I come from a History and biology background.
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Maths_Lover
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(Original post by tennantgirl)
Hello all
I have a quick question on a homework sheet which I am a little stuck on how to work it out- trust me I have tried!
so it's...

1) you are given the identity x² -12x+10= (x=a)² +b
work out the values of a and b

2) The diameter of the base of the cone is xcm. The height of the cone is also xcm. The volume of the cone is Vcm³.
Find the formula for x in terms of V and x.

So far I have got V=1/3 ? 1/2x²x ...but I don't know what to do now.

Any help would be really appreciated, as I've been stuck on these questions for ages. Thank you!
1) x^2-12x+10 = (x-a)^2 +b
First multiply out the brackets on the RHS:

x^2-12x+10 = x^2 -2ax + a^2 +b

Comparing coefficients means that:

-12 = -2a
a=6

a^2 +b = 10
but a=6, so:
6^2 +b = 10
36 +b = 10
b = -26

2) Volume of a cone is: 1/3pir^2h

diameter of base = x
So the radius is x/2
height = x

So substituting into the above formula gives:

V = 1/3pi((x/2)^2)x

Simplify the RHS:

V = 1/3pi(x^2/4)x
V = 1/3pi(x^3)/4

Next, we rearrange the formula above to give x in terms of V and x:

V = 1/3pi(x^3)/4
V= pi/3(x^3)/4

mutiply both sides by 12 to get rid of the two fractions on the RHS:

12V = 4pi3x^3

Divide both sides by 4pi:

3x^3 = 12V/4pi
3x^3 = 3V/pi

Divide both sides by 3:

x^3 = 3V/3pi
x^3 = V/pi

Finally, to get rid of the x cubed on the LHS, take the cube root of both sides:

x = cuberoot(V/pi)

Sorry for the length of this, but I hope it helps
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tennantgirl
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(Original post by Maths_Lover)
1) x^2-12x+10 = (x-a)^2 +b
First multiply out the brackets on the RHS:

x^2-12x+10 = x^2 -2ax + a^2 +b

Comparing coefficients means that:

-12 = -2a
a=6

a^2 +b = 10
but a=6, so:
6^2 +b = 10
36 +b = 10
b = -26

2) Volume of a cone is: 1/3pir^2h

diameter of base = x
So the radius is x/2
height = x

So substituting into the above formula gives:

V = 1/3pi((x/2)^2)x

Simplify the RHS:

V = 1/3pi(x^2/4)x
V = 1/3pi(x^3)/4

Next, we rearrange the formula above to give x in terms of V and x:

V = 1/3pi(x^3)/4
V= pi/3(x^3)/4

mutiply both sides by 12 to get rid of the two fractions on the RHS:

12V = 4pi3x^3

Divide both sides by 4pi:

3x^3 = 12V/4pi
3x^3 = 3V/pi

Divide both sides by 3:

x^3 = 3V/3pi
x^3 = V/pi

Finally, to get rid of the x cubed on the LHS, take the cube root of both sides:

x = cuberoot(V/pi)

Sorry for the length of this, but I hope it helps
thank you so much
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Maths_Lover
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(Original post by tennantgirl)
thank you so much
You are welcome
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