# OCR MEI C2 june 2017

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What did every one get for the log questions where you have to find w?

Also for integration question on the first page. When the second part didn't have the limit, did you put +c at the end?

How did you find the paper?

Also for integration question on the first page. When the second part didn't have the limit, did you put +c at the end?

How did you find the paper?

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#2

(Original post by

What did every one get for the log questions where you have to find w?

Also for integration question on the first page. When the second part didn't have the limit, did you put +c at the end?

How did you find the paper?

**Unefleur**)What did every one get for the log questions where you have to find w?

Also for integration question on the first page. When the second part didn't have the limit, did you put +c at the end?

How did you find the paper?

And yes you have to put +c for one mark

Overall I thought it was ok, although time was a big issue with section B!

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(Original post by

I found the log w question a bit weird, I got something like w=a^3 + 3x^4? Can't really remember though!

And yes you have to put +c for one mark

Overall I thought it was ok, although time was a big issue with section B!

**charlottesacha99**)I found the log w question a bit weird, I got something like w=a^3 + 3x^4? Can't really remember though!

And yes you have to put +c for one mark

Overall I thought it was ok, although time was a big issue with section B!

I though section B was good compared to section A. For the surface area question i got r as 3.99 and surface area as 101. For the car park meadow and pond question I got 93.7% or 94.7% can't remember which one though. And for the last question I got n as 25

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#4

**Unefleur**)

What did every one get for the log questions where you have to find w?

Also for integration question on the first page. When the second part didn't have the limit, did you put +c at the end?

How did you find the paper?

That's what I got

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#6

i think it was logw = 3 + logx^5 - log2x + log6

i got

3a^3x^4

i got

3a^3x^4

(Original post by

Do you remember the exact question for the log question?

**Unefleur**)Do you remember the exact question for the log question?

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#7

(Original post by

I didn't get that for the log quesions. Think that's the correct answer as all the other people I talked to got that.

I though section B was good compared to section A. For the surface area question i got r as 3.99 and surface area as 101. For the car park meadow and pond question I got 93.7% or 94.7% can't remember which one though. And for the last question I got n as 25

**Unefleur**)I didn't get that for the log quesions. Think that's the correct answer as all the other people I talked to got that.

I though section B was good compared to section A. For the surface area question i got r as 3.99 and surface area as 101. For the car park meadow and pond question I got 93.7% or 94.7% can't remember which one though. And for the last question I got n as 25

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#9

(Original post by

if you put r as 3.99, back into the equation for A you should have ended up with 300.49... which should be rounded to 301 or 300.5

**Levisual360**)if you put r as 3.99, back into the equation for A you should have ended up with 300.49... which should be rounded to 301 or 300.5

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#10

(Original post by

Would it be okay if you rounded 3.99 to 4.0 but left the minimum point as a surd

**Aminah53**)Would it be okay if you rounded 3.99 to 4.0 but left the minimum point as a surd

if you just straight worked out r to be 4, then im not sure if you would get the mark.

As for the surd form, im not too sure, it depends on the examiner. I think as long as youve shown ur substitution of r back into the equation for A, and simplified that should be fine.

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#11

How do you find the minimum radius r?? I took the second differential and made it equal to 0 and got r as 5 and that was the wrong answer

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#13

This is what I can remember, random order :/

1)

a) Integral 5 to 1: 4xdx = [2x^2] = 50 โ 2 = 48.

b) Integral: 6x^1/2 = 4x^3/2 + c

2)

a) (log0.2)-(log0.1)/(0.2-0.1) = 3.01

b) Plot c in between a and b

3)

Log

Log

Log

Log

Log

W = 3a^3x^4

4) Y=2x^3. Normal at x = 2. y = 16. Dy/dx = 6x^2. Dy/dx = 24. Normal = -1/24. Equation: x + 24y = 386.

5)

a) Cosine rule. a^2 = 32^2 + 15^2 - (2 x 32 x 15 x cos116). AE = 40.86โฆ.

b)

Perp. Distance from AE to D.

Area ADE = 1/2 absinC = 1/2 x 32 x 15 x sin116 = 215.7.

Area ADE = b x h x 1/2.

215.7... = 40.86... x h x 1/2

h = 10.55โฆ > 10. Therefore pond lies in triangle.

c)

Area of pond = 116/360 x Pi x 10^2 = 101.22โฆ

Area of ADE = 215.7โฆ

Area of meadow = 114.48โฆ

d)

Angle C = 360 - 90 - 90 - 116 = 64.

Use that to work out total length of base of trapezium = 70 something.

Area of trapezium = 1/2 x (32 + 70 something) x 80 = 4120.74โฆ

Area of car park = trapezium โ ADE = 3905.033โฆ

90% of trapezium = 3708.66โฆ

3905>3708 so carpark takes up more than 90% of field.

6)

6cosx^2 = 5 - sinx

6(1-sinx^2) = 5 - sinx

6 - 6sinx^2 = 5 - sinx

6sinx^2 - sinx - 1 = 0

(3sinx + 1)(2sinx - 1) = 0

sinx = -1/3 or 1/2

x = 3.48, 5.94, 1/6PI, 5/6Pi

7)

a)

Asif = 30000 + (n-1)1000. Bettina = 25000 x 1.05^n-1.

10

11

b)

Total after 17 years.

Asif = 646000. Bettina = 646000 (to the nearest hundred). The same

c)

Total > ยฃM.

25000(1.05^n โ 1)/1.05-1 > M

25000(1.05^n โ 1)/0.05 > M

25000(1.05^n โ 1) > 0.05M

500000(1.05^n โ 1) > M

(500000 x 1.05^n) โ 500000 > M

500000 x 1.05^n > M + 500000

1.05^n > (M + 500000)/500000

log1.05^n > log (M + 500000) โ log500000

nlog1.05 > log (M + 500000) โ log500000

n > (log (M + 500000) โ log500000)/log1.05

n = 25 or 26 :/

8)

a)

Stretch parallel to y-axis SF 2.

b)

Translation (3 0)

9)

Curve goes through (2,10)

Dy/dx = 12x^3 - 7

Equation of curve: y = 3x^4 -7x + c

c = -24

y = 3x^4 -7x - 24

10)

a)

V = 400

400 = Pi r^2 h

h = 400/Pir^2

A = 2Pir^2 + 2Pirh

A = 2Pir^2 + 800/r

b)

dA/dr = 4Pir -800/r^2

d2A/dr2 = 4Pi +1600/r^3

c)

Min value of r = 3.992..

d2A/dr2 = 37.6... > 0 Therefore minimum.

A with this value = 300.53.. = 301

11)

a) Sketch graph y = 2^x. Goes through (0,1)

b) ?

12)

a)

Sum of (3r + 2) from 1 to 5 = 5 + 8 + 11 + 14 + 17 = 55

b)

AP First term = 4.2. Sixth term = 1.8.

4.2 + 5d = 1.8

5d = -2.4

d = -0.48

1)

a) Integral 5 to 1: 4xdx = [2x^2] = 50 โ 2 = 48.

b) Integral: 6x^1/2 = 4x^3/2 + c

2)

a) (log0.2)-(log0.1)/(0.2-0.1) = 3.01

b) Plot c in between a and b

3)

Log

_{a}w = 3 + log_{a}x^5 + log_{a}6 โ log_{a}2xLog

_{a}w = log_{a}a^3 + log_{a}x^5 + log_{a}6 โ log_{a}2xLog

_{a}w = log_{a}(6a^3x^5) โ log_{a}2xLog

_{a}w = log_{a}(6a^3x^5/2x)Log

_{a}w = log_{a}(3a^3x^4)W = 3a^3x^4

4) Y=2x^3. Normal at x = 2. y = 16. Dy/dx = 6x^2. Dy/dx = 24. Normal = -1/24. Equation: x + 24y = 386.

5)

a) Cosine rule. a^2 = 32^2 + 15^2 - (2 x 32 x 15 x cos116). AE = 40.86โฆ.

b)

Perp. Distance from AE to D.

Area ADE = 1/2 absinC = 1/2 x 32 x 15 x sin116 = 215.7.

Area ADE = b x h x 1/2.

215.7... = 40.86... x h x 1/2

h = 10.55โฆ > 10. Therefore pond lies in triangle.

c)

Area of pond = 116/360 x Pi x 10^2 = 101.22โฆ

Area of ADE = 215.7โฆ

Area of meadow = 114.48โฆ

d)

Angle C = 360 - 90 - 90 - 116 = 64.

Use that to work out total length of base of trapezium = 70 something.

Area of trapezium = 1/2 x (32 + 70 something) x 80 = 4120.74โฆ

Area of car park = trapezium โ ADE = 3905.033โฆ

90% of trapezium = 3708.66โฆ

3905>3708 so carpark takes up more than 90% of field.

6)

6cosx^2 = 5 - sinx

6(1-sinx^2) = 5 - sinx

6 - 6sinx^2 = 5 - sinx

6sinx^2 - sinx - 1 = 0

(3sinx + 1)(2sinx - 1) = 0

sinx = -1/3 or 1/2

x = 3.48, 5.94, 1/6PI, 5/6Pi

7)

a)

Asif = 30000 + (n-1)1000. Bettina = 25000 x 1.05^n-1.

10

^{th}year โ Asif = ยฃ39000. Bettina = ยฃ38783. Asif had more.11

^{th}year โ Asif = ยฃ40000. Bettina = ยฃ40722. Bettina had moreb)

Total after 17 years.

Asif = 646000. Bettina = 646000 (to the nearest hundred). The same

c)

Total > ยฃM.

25000(1.05^n โ 1)/1.05-1 > M

25000(1.05^n โ 1)/0.05 > M

25000(1.05^n โ 1) > 0.05M

500000(1.05^n โ 1) > M

(500000 x 1.05^n) โ 500000 > M

500000 x 1.05^n > M + 500000

1.05^n > (M + 500000)/500000

log1.05^n > log (M + 500000) โ log500000

nlog1.05 > log (M + 500000) โ log500000

n > (log (M + 500000) โ log500000)/log1.05

n = 25 or 26 :/

8)

a)

Stretch parallel to y-axis SF 2.

b)

Translation (3 0)

9)

Curve goes through (2,10)

Dy/dx = 12x^3 - 7

Equation of curve: y = 3x^4 -7x + c

c = -24

y = 3x^4 -7x - 24

10)

a)

V = 400

400 = Pi r^2 h

h = 400/Pir^2

A = 2Pir^2 + 2Pirh

A = 2Pir^2 + 800/r

b)

dA/dr = 4Pir -800/r^2

d2A/dr2 = 4Pi +1600/r^3

c)

Min value of r = 3.992..

d2A/dr2 = 37.6... > 0 Therefore minimum.

A with this value = 300.53.. = 301

11)

a) Sketch graph y = 2^x. Goes through (0,1)

b) ?

12)

a)

Sum of (3r + 2) from 1 to 5 = 5 + 8 + 11 + 14 + 17 = 55

b)

AP First term = 4.2. Sixth term = 1.8.

4.2 + 5d = 1.8

5d = -2.4

d = -0.48

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#14

[QUOTE=gemma.louise;71942372]w = 1/12 aยณ x^4

That's what I got[/QUOTE

Nicee I got that as well

That's what I got[/QUOTE

Nicee I got that as well

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#15

(Original post by

What did everyone get for the car park field percentage?

**Eldronyx**)What did everyone get for the car park field percentage?

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#16

(Original post by

This is what I can remember, random order :/

1)

a) Integral 5 to 1: 4xdx = [2x^2] = 50 โ 2 = 48.

b) Integral: 6x^1/2 = 4x^3/2 + c

2)

a) (log0.2)-(log0.1)/(0.2-0.1) = 3.01

b) Plot c in between a and b

3)

Log

Log

Log

Log

Log

W = 3a^3x^4

4) Y=2x^3. Normal at x = 2. y = 16. Dy/dx = 6x^2. Dy/dx = 24. Normal = -1/24. Equation: x + 24y = 386.

5)

a) Cosine rule. a^2 = 32^2 + 15^2 - (2 x 32 x 15 x cos116). AE = 40.86โฆ.

b)

Perp. Distance from AE to D.

Area ADE = 1/2 absinC = 1/2 x 32 x 15 x sin116 = 215.7.

Area ADE = b x h x 1/2.

215.7... = 40.86... x h x 1/2

h = 10.55โฆ > 10. Therefore pond lies in triangle.

c)

Area of pond = 116/360 x Pi x 10^2 = 101.22โฆ

Area of ADE = 215.7โฆ

Area of meadow = 114.48โฆ

d)

Angle C = 360 - 90 - 90 - 116 = 64.

Use that to work out total length of base of trapezium = 70 something.

Area of trapezium = 1/2 x (32 + 70 something) x 80 = 4120.74โฆ

Area of car park = trapezium โ ADE = 3905.033โฆ

90% of trapezium = 3708.66โฆ

3905>3708 so carpark takes up more than 90% of field.

6)

6cosx^2 = 5 - sinx

6(1-sinx^2) = 5 - sinx

6 - 6sinx^2 = 5 - sinx

6sinx^2 - sinx - 1 = 0

(3sinx + 1)(2sinx - 1) = 0

sinx = -1/3 or 1/2

x = 3.48, 5.94, 1/6PI, 5/6Pi

7)

a)

Asif = 30000 + (n-1)1000. Bettina = 25000 x 1.05^n-1.

10

11

b)

Total after 17 years.

Asif = 646000. Bettina = 646000 (to the nearest hundred). The same

c)

Total > ยฃM.

25000(1.05^n โ 1)/1.05-1 > M

25000(1.05^n โ 1)/0.05 > M

25000(1.05^n โ 1) > 0.05M

500000(1.05^n โ 1) > M

(500000 x 1.05^n) โ 500000 > M

500000 x 1.05^n > M + 500000

1.05^n > (M + 500000)/500000

log1.05^n > log (M + 500000) โ log500000

nlog1.05 > log (M + 500000) โ log500000

n > (log (M + 500000) โ log500000)/log1.05

n = 25 or 26 :/

8)

a)

Stretch parallel to y-axis SF 2.

b)

Translation (3 0)

9)

Curve goes through (2,10)

Dy/dx = ? - 7.

Equation of curve: y = ? -7x + c

c = ?

10)

a)

V = 400

400 = Pi r^2 h

h = 400/Pir^2

A = 2Pir^2 + 2Pirh

A = 2Pir^2 + 800/r

b)

dA/dr = 4Pir -800/r^2

d2A/dr2 = 4Pi +1600/r^3

c)

Min value of r = 3.992..

d2A/dr2 = 37.6... > 0 Therefore minimum.

A with this value = 300.53.. = 301

11)

a) Sketch graph y = 2^x. Goes through (0,1)

b) ?

**wmirza**)This is what I can remember, random order :/

1)

a) Integral 5 to 1: 4xdx = [2x^2] = 50 โ 2 = 48.

b) Integral: 6x^1/2 = 4x^3/2 + c

2)

a) (log0.2)-(log0.1)/(0.2-0.1) = 3.01

b) Plot c in between a and b

3)

Log

_{a}w = 3 + log_{a}x^5 + log_{a}6 โ log_{a}2xLog

_{a}w = log_{a}a^3 + log_{a}x^5 + log_{a}6 โ log_{a}2xLog

_{a}w = log_{a}(6a^3x^5) โ log_{a}2xLog

_{a}w = log_{a}(6a^3x^5/2x)Log

_{a}w = log_{a}(3a^3x^4)W = 3a^3x^4

4) Y=2x^3. Normal at x = 2. y = 16. Dy/dx = 6x^2. Dy/dx = 24. Normal = -1/24. Equation: x + 24y = 386.

5)

a) Cosine rule. a^2 = 32^2 + 15^2 - (2 x 32 x 15 x cos116). AE = 40.86โฆ.

b)

Perp. Distance from AE to D.

Area ADE = 1/2 absinC = 1/2 x 32 x 15 x sin116 = 215.7.

Area ADE = b x h x 1/2.

215.7... = 40.86... x h x 1/2

h = 10.55โฆ > 10. Therefore pond lies in triangle.

c)

Area of pond = 116/360 x Pi x 10^2 = 101.22โฆ

Area of ADE = 215.7โฆ

Area of meadow = 114.48โฆ

d)

Angle C = 360 - 90 - 90 - 116 = 64.

Use that to work out total length of base of trapezium = 70 something.

Area of trapezium = 1/2 x (32 + 70 something) x 80 = 4120.74โฆ

Area of car park = trapezium โ ADE = 3905.033โฆ

90% of trapezium = 3708.66โฆ

3905>3708 so carpark takes up more than 90% of field.

6)

6cosx^2 = 5 - sinx

6(1-sinx^2) = 5 - sinx

6 - 6sinx^2 = 5 - sinx

6sinx^2 - sinx - 1 = 0

(3sinx + 1)(2sinx - 1) = 0

sinx = -1/3 or 1/2

x = 3.48, 5.94, 1/6PI, 5/6Pi

7)

a)

Asif = 30000 + (n-1)1000. Bettina = 25000 x 1.05^n-1.

10

^{th}year โ Asif = ยฃ39000. Bettina = ยฃ38783. Asif had more.11

^{th}year โ Asif = ยฃ40000. Bettina = ยฃ40722. Bettina had moreb)

Total after 17 years.

Asif = 646000. Bettina = 646000 (to the nearest hundred). The same

c)

Total > ยฃM.

25000(1.05^n โ 1)/1.05-1 > M

25000(1.05^n โ 1)/0.05 > M

25000(1.05^n โ 1) > 0.05M

500000(1.05^n โ 1) > M

(500000 x 1.05^n) โ 500000 > M

500000 x 1.05^n > M + 500000

1.05^n > (M + 500000)/500000

log1.05^n > log (M + 500000) โ log500000

nlog1.05 > log (M + 500000) โ log500000

n > (log (M + 500000) โ log500000)/log1.05

n = 25 or 26 :/

8)

a)

Stretch parallel to y-axis SF 2.

b)

Translation (3 0)

9)

Curve goes through (2,10)

Dy/dx = ? - 7.

Equation of curve: y = ? -7x + c

c = ?

10)

a)

V = 400

400 = Pi r^2 h

h = 400/Pir^2

A = 2Pir^2 + 2Pirh

A = 2Pir^2 + 800/r

b)

dA/dr = 4Pir -800/r^2

d2A/dr2 = 4Pi +1600/r^3

c)

Min value of r = 3.992..

d2A/dr2 = 37.6... > 0 Therefore minimum.

A with this value = 300.53.. = 301

11)

a) Sketch graph y = 2^x. Goes through (0,1)

b) ?

dy/dx = 12x^3 - 7, given that the curve goes through (2,10)

so c should be =-24

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#17

Cheers

(Original post by

For the integral question it was asking to integrate

dy/dx = 12x^3 - 7, given that the curve goes through (2,10)

so c should be =-24

**Levisual360**)For the integral question it was asking to integrate

dy/dx = 12x^3 - 7, given that the curve goes through (2,10)

so c should be =-24

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**wmirza**)

This is what I can remember, random order :/

1)

a) Integral 5 to 1: 4xdx = [2x^2] = 50 โ 2 = 48.

b) Integral: 6x^1/2 = 4x^3/2 + c

2)

a) (log0.2)-(log0.1)/(0.2-0.1) = 3.01

b) Plot c in between a and b

3)

Log

_{a}w = 3 + log

_{a}x^5 + log

_{a}6 โ log

_{a}2x

Log

_{a}w = log

_{a}a^3 + log

_{a}x^5 + log

_{a}6 โ log

_{a}2x

Log

_{a}w = log

_{a}(6a^3x^5) โ log

_{a}2x

Log

_{a}w = log

_{a}(6a^3x^5/2x)

Log

_{a}w = log

_{a}(3a^3x^4)

W = 3a^3x^4

4) Y=2x^3. Normal at x = 2. y = 16. Dy/dx = 6x^2. Dy/dx = 24. Normal = -1/24. Equation: x + 24y = 386.

5)

a) Cosine rule. a^2 = 32^2 + 15^2 - (2 x 32 x 15 x cos116). AE = 40.86โฆ.

b)

Perp. Distance from AE to D.

Area ADE = 1/2 absinC = 1/2 x 32 x 15 x sin116 = 215.7.

Area ADE = b x h x 1/2.

215.7... = 40.86... x h x 1/2

h = 10.55โฆ > 10. Therefore pond lies in triangle.

c)

Area of pond = 116/360 x Pi x 10^2 = 101.22โฆ

Area of ADE = 215.7โฆ

Area of meadow = 114.48โฆ

d)

Angle C = 360 - 90 - 90 - 116 = 64.

Use that to work out total length of base of trapezium = 70 something.

Area of trapezium = 1/2 x (32 + 70 something) x 80 = 4120.74โฆ

Area of car park = trapezium โ ADE = 3905.033โฆ

90% of trapezium = 3708.66โฆ

3905>3708 so carpark takes up more than 90% of field.

6)

6cosx^2 = 5 - sinx

6(1-sinx^2) = 5 - sinx

6 - 6sinx^2 = 5 - sinx

6sinx^2 - sinx - 1 = 0

(3sinx + 1)(2sinx - 1) = 0

sinx = -1/3 or 1/2

x = 3.48, 5.94, 1/6PI, 5/6Pi

7)

a)

Asif = 30000 + (n-1)1000. Bettina = 25000 x 1.05^n-1.

10

^{th}year โ Asif = ยฃ39000. Bettina = ยฃ38783. Asif had more.

11

^{th}year โ Asif = ยฃ40000. Bettina = ยฃ40722. Bettina had more

b)

Total after 17 years.

Asif = 646000. Bettina = 646000 (to the nearest hundred). The same

c)

Total > ยฃM.

25000(1.05^n โ 1)/1.05-1 > M

25000(1.05^n โ 1)/0.05 > M

25000(1.05^n โ 1) > 0.05M

500000(1.05^n โ 1) > M

(500000 x 1.05^n) โ 500000 > M

500000 x 1.05^n > M + 500000

1.05^n > (M + 500000)/500000

log1.05^n > log (M + 500000) โ log500000

nlog1.05 > log (M + 500000) โ log500000

n > (log (M + 500000) โ log500000)/log1.05

n = 25 or 26 :/

8)

a)

Stretch parallel to y-axis SF 2.

b)

Translation (3 0)

9)

Curve goes through (2,10)

Dy/dx = ? - 7.

Equation of curve: y = ? -7x + c

c = ?

10)

a)

V = 400

400 = Pi r^2 h

h = 400/Pir^2

A = 2Pir^2 + 2Pirh

A = 2Pir^2 + 800/r

b)

dA/dr = 4Pir -800/r^2

d2A/dr2 = 4Pi +1600/r^3

c)

Min value of r = 3.992..

d2A/dr2 = 37.6... > 0 Therefore minimum.

A with this value = 300.53.. = 301

11)

a) Sketch graph y = 2^x. Goes through (0,1)

b) ?

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In 5)b) answer you've given did we have to compare the height to something? Did I not read the questions properly omg๐ฑ

**Unefleur**)In 5)b) answer you've given did we have to compare the height to something? Did I not read the questions properly omg๐ฑ

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So you needed to find the perpendicular distance from AE to D, which is the height of the triangle ADE, and AE being the base

**wmirza**)So you needed to find the perpendicular distance from AE to D, which is the height of the triangle ADE, and AE being the base

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