pls help with C1 questions! (pleeease!!!) Watch

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*girlie*
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6) The straight line l has the equation x-2y=12 and meets the coordinate axes at the points A and B.
Find the distance of the mid-point of AB from the origin, giving your answer in the form: k sqrt 5.

7) a) Given that y=2^x, find expressions in terms of y for
i) 2^x+2
ii) 2^3-x

c) Show that using the substitution y=2^x, the equation 2^x+2 + 2^3-x =33 can be rewritten as 4y^2 – 33y + 8 =0.
d) Hence solve the equation 2^x+2 + 2^3-x =33

9) The second and fifth terms of an arithmetic series are 26 and 41 respectively.
a) Show that the common difference of the series is 5.
b) Find the 12th term of the series.

Another arithmetic series has first term –12 and common difference 7.
Given that the sums of the first n terms of these two series are equal,
c) find the value of n.
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thefish_uk
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(Original post by *girlie*)
6) The straight line l has the equation x-2y=12 and meets the coordinate axes at the points A and B.
Find the distance of the mid-point of AB from the origin, giving your answer in the form: k sqrt 5.
Right....

Say at A it crosses the y axis so x = 0 so then - 2y = 12 so y = -6

And at B it crosses the x axis so y = 0 so then x = 12

Now we need the mid point of the line from (0, -6) to (12, 0), which is simply the averages of the x and y co-ordinates, so that's (6, -3)

If we draw a line from the origin to (6, -3) that'll simply go right 6 and down 3 so its length is sqrt(6^2 + 3^2) by Pythagoras' theorem, which is sqrt(45) which is sqrt (3*3*5) which is 3 sqrt 5 and theres the answer
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thefish_uk
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(Original post by *girlie*)
7) a) Given that y=2^x, find expressions in terms of y for
i) 2^x+2
ii) 2^3-x
I take it you mean

i) 2^(x+2)
ii) 2^(3-x)

So for i)
2^(x+2) is the same as 2^x * x^2 so thats y * 2^2 which is 4y

For ii)
2^(3-x) is the same as 2^3 divided by 2^x which is 8/y

(Original post by *girlie*)
c) Show that using the substitution y=2^x, the equation 2^(x+2) + 2^(3-x) =33 can be rewritten as 4y^2 – 33y + 8 =0.
d) Hence solve the equation 2^x+2 + 2^3-x =33
c) so substituting what we worked out earlier...

4y + (8/y) = 33
4y^2 + 8 = 33y
4y^2 - 33y + 8 = 0

d) just use quadratic formula
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thefish_uk
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(Original post by *girlie*)
9) The second and fifth terms of an arithmetic series are 26 and 41 respectively.
a) Show that the common difference of the series is 5.
b) Find the 12th term of the series.
Second term = 26 = a + d
Fifth term = 41 = a + 4d

Taking the second term from the fifth term:

15 = 3d
d = 5

That's our answer to a)

So for b)

We need to find what a is, that's 26-5 so a = 21

So 12th term = 21+11(5) = 21 + 55 = 76

Another arithmetic series has first term –12 and common difference 7.
Given that the sums of the first n terms of these two series are equal,
c) find the value of n.
Hmmm...

Sum of first n terms = n[2a+(n-1)d)]/2
So sum of first n terms of first one:
= n[2*21 + (n-1)5)/2
= n[42 + 5n - 5]/2
= n(37+5n)/2
= n(18.5 + 2.5n)
= 2.5n^2 + 18.5n

Sum of first n terms of second one:
= n[2*-12 + (n-1)7]/2
= n[-12 + 3.5n - 3.5]
= n(3.5n - 15.5)
= 3.5n^2 - 15.5n

And we're told that they're equal...
2.5n^2 + 18.5n = 3.5n^2 - 15.5n
n^2 - 34n=0
n(n-34) = 0

So n = 34 or 0 so the sum of the first 34 terms are the same
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Christophicus
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9) The second and fifth terms of an arithmetic series are 26 and 41 respectively.
a) Show that the common difference of the series is 5.
b) Find the 12th term of the series.

Another arithmetic series has first term –12 and common difference 7.
Given that the sums of the first n terms of these two series are equal,
c) find the value of n.


a) Use un = a + (n-1)d
u2 = a + d = 26
u5 = a + 4d = 41

3d = 15
d = 5

b)
a + d = 26
a = 26 - 5 = 21

u12 = a + 5(12-1)
u12 = 21 + 55
u12 = 76

c)
1st arithmetic series
Sn = n/2(42 + 5(n-1))

2nd arithmetic series
Sn = n/2(-24 + 7(n-1)

Two series are equal

n/2(42 + 5n - 5) = n/2(-24 + 7n - 7)
n/2 cancels

37 + 5n = -31 + 7n
68 = 2n
n = 34
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