Question

Help! Picture attached. Showing working out thanks.

I need help with Q1.b and 2 from the document and
Q4,5 and 7 from the picture

Answer 1

so where should I start?

Answer 2

I did #10, #13 and #15 on another post, did you see it?

Answer 3

with the worded questions in the document one

Answer 4

yepyep. thanks.

Answer 5

so start with Q1 a)?

Answer 6

well i did 1.a but I'm having trouble with 1.b

Answer 7

alright, are you comfortable with calculus?

Answer 8

some of it I believe.

Answer 9

how about derivatives?

Answer 10

nope.

Answer 11

alright, just thought I'd throw that out there

Answer 12

well i haven't learnt it yet and all i know in regards to it is dy/dx

Answer 13

alright one sec while I think of an alternative

Answer 14

kk thanks.

Answer 15

oh after plotting it, it turns out that there's some nice symmetry going on that I didn't think about

The figure is symmetrical with respect to the y-axis.

So we can reflect y = sqrt(15)*x over the y axis to get


y = sqrt(15)*x

y = sqrt(15)*(-x) ... reflect over the y-axis by replacing 'x' with '-x'

y = -sqrt(15)*x


So the only other equation that is tangent to the circle and that passes through (0,0) is y = -sqrt(15)*x

Answer 16

where to next?

Answer 17

um i still don't quite get how your working out.

Answer 18

I simply replaced x with -x and simplified

Answer 19

I'm able to do this because the figure is symmetrical with respect to the y-axis (ie you can fold the circle over the y-axis and it will line up perfectly)

Answer 20

kk but how does y = sqrt(15)*(-x) turn into y = -sqrt(15)*x?

Answer 21

oh wait u just times right?

Answer 22

yes, I'm just using the commutativity of multiplication (ie I'm moving around that negative 1)

Answer 23

kk

Answer 24

can u help me with question 2 on the same document?

Answer 25

2 a)? 2 b)? or some other part?

Answer 26

2.b), 2.c.iii) and 2d onwards.

Answer 27

2 b)

domain = set of all real numbers but x can't equal -2 (to avoid dividing by zero)

range = set of all real numbers but y can't equal 3 (this is the horizontal asymptote)

Answer 28

2 c)


i)

y = 1/(x+2) + 3

y = 1/(8+2) + 3

y = 1/10 + 3

y = 31/10

So y = 31/10 when x = 8

ii)

y = 1/(x+2) + 3

8 = 1/(x+2) + 3

8-3 = 1/(x+2)

5 = 1/(x+2)

5(x+2) = 1

x+2 = 1/5

x = 1/5-2

x = -9/5

So x = -9/5 when y = 8


iii)

The right side edge of piece one is 8 - 31/10 = 49/10 = 4.9 units

The top edge is 8+9/5 = 49/5 = 9.8 units

So the total of the straight edges is 4.9+9.8 = 14.7 units

Answer 29

sorry if i didn't say it clearly b4 but u only had to do c.iii)

Answer 30

out of the c part

Answer 31

2 d)

To reflect over the line y = x, you basically swap x and y. Then you solve for y


First simplify

y = 1/(x+2) + 3

y = 1/(x+2) + 3(x+2)/(x+2)

y = 1/(x+2) + (3x+6)/(x+2)

y = (1+3x+6)/(x+2)

y = (3x+7)/(x+2)


Then swap x and y and then solve for y


y = (3x+7)/(x+2)

x = (3y+7)/(y+2)

x(y+2) = 3y+7

xy+2x = 3y+7

xy+2x-3y= 7

xy-3y= -2x+7

y(x-3)= -2x+7

y= (-2x+7)/(x-3)


So after reflecting y = (3x+7)/(x+2) over the line y = x, we get y= (-2x+7)/(x-3)

Answer 32

for 2 e), you just want part iii only?

Answer 33

sorry i dced because the internet screwed up.

Answer 34

thats ok, so do you need all of 2 e)?

Answer 35

yes thanks

Answer 36

well I think ur d is wrong cause i have solutions.

Answer 37

what solutions do you have?

Answer 38

i didn't write the solution btw

Answer 39

hmm guess they don't want you to solve for y, so the answer is just


x = 1/(y+2) + 3

Answer 40

i guess that simplifies things greatly

Answer 41

the answer is y= 1/x-3 -2

Answer 42

oh so they wanted to solve for y, alright one sec

Answer 43

yeh. could u explain what it means by "reflecting the edges of the pieces in Diagram One in the line y=x"? and "Find the equation which defines the curves in this differently configured puzzle?" I'm still confused.

Answer 44

2 d)



To reflect any curve over y = x, you just swap x and y. Afterwards you solve for y (this is basically finding the inverse)


y = 1/(x+2) + 3

y = 1/(x+2) + 3(x+2)/(x+2)

y = 1/(x+2) + (3x+6)/(x+2)

y = (1+3x+6)/(x+2)

y = (3x+7)/(x+2)


Then swap x and y and then solve for y


y = (3x+7)/(x+2)

x = (3y+7)/(y+2)

x(y+2) = 3y+7

xy+2x = 3y+7

xy+2x-3y= 7

xy-3y= -2x+7

y(x-3)= -2x+7

y= (-2x+7)/(x-3)


Now simplify


y= (-2x+7)/(x-3)

y= (-2x+6+1)/(x-3)

y= (-2(x-3)+1)/(x-3)

y= 1/(x-3) - 2(x-3)/(x-3)

y= 1/(x-3) - 2

Answer 45

wow so many steps, i think it's better to follow the solution steps for d since it's less steps.

Answer 46

they swapped x and y at the very start

Answer 47

and just rearranged.

Answer 48

alright, so why did I do that one exactly?

Answer 49

i wanted to understand it properly and see if you had a better way to do it.

Answer 50

oh, i see

Answer 51

I'll finish up with 2 e) then I have to get going

Answer 52

kk, thanks.

Answer 53

2e)

i

y-intercept --- plug in x = 0 and solve for y

y = -50/((x+10)^2) - 4

y = -50/((0+10)^2) - 4

y = -50/((10)^2) - 4

y = -50/(100) - 4

y = -1/2 - 4

y = -9/2

So the y-intercept is (0, -9/2)


ii

Plug in x = 8 and evaluate to find y


y = -50/((x+10)^2) - 4

y = -50/((8+10)^2) - 4

y = -50/((18)^2) - 4

y = -50/(324) - 4

y = -25/162 - 4

y = -673/162


So when x = 8, the value of y is y = -673/162


iii

Plug in y = -8 and then solve for x

y = -50/((x+10)^2) - 4

-8 = -50/((x+10)^2) - 4

-8+4 = -50/((x+10)^2)

-4 = -50/((x+10)^2)

-1/4 = -((x+10)^2)/50

50/4 = (x+10)^2

25/2 = (x+10)^2

(x+10)^2 = 25/2

x+10 = sqrt(25/2) or x+10 = -sqrt(25/2)

x+10 = (5/2)*sqrt(2) or x+10 = -(5/2)*sqrt(2)

x+10 = (5/2)*sqrt(2) or x+10 = -(5/2)*sqrt(2)

x = -10+(5/2)*sqrt(2) or x = -10-(5/2)*sqrt(2)

So when y = -8, the values of x are x = -10+(5/2)*sqrt(2) or x = -10-(5/2)*sqrt(2)

Answer 54

ok so basically it's just substituting and solving. for part e.

Answer 55

that's exactly right

Answer 56

sqrt(25/2)


sqrt(25)/sqrt(2)


5/sqrt(2)


(5/sqrt(2))*(sqrt(2)/sqrt(2))


(5*sqrt(2))/(sqrt(2)*sqrt(2))


(5*sqrt(2))/(sqrt(2*2))


(5*sqrt(2))/(sqrt(4))


(5*sqrt(2))/2


(5/2)*sqrt(2)



So sqrt(25/2) = (5/2)*sqrt(2)


So that's where the sqrt(2) comes from

Answer 57

oh okay so u times top and bottom by sqrt 2 to get rid of the sqrt2 denominator.

Answer 58

that's wat u were doing.

Answer 59

yes, i'm rationalizing the denominator

Answer 60

i thought u could just leave it as 5 over sqrt2

Answer 61

you can, but most books will rationalize the denominator

Answer 62

kk understood. thanks

Answer 63

that's great, you're welcome

Answer 64

ok I'm going to close this post and create a new one.

Answer 65

alright

Answer 66

by any chance can u still access the posts that u closed?

Answer 67

i think so? I'm not too sure

Answer 68

kk thanks anyway.

Answer 69

alright