Question

help with this assignment please ! my equation is y = 3x + 6.

Answer 1

@mathstudent55

Answer 2

@whpalmer4

Answer 3

So your path is \(y = 3x+6\)? And you need to find where that intersects with \(y^2+x^2=40000\)?

Why don't you substitute the right hand side of your equation into the other equation wherever you see \(y\)? That should give you something you can solve.

Answer 4

yes

Answer 5

so \[3x + 6^2\]

Answer 6

\[+ x^2 = 40,000\]

Answer 7

No, you have to substitute like this:

\[y^2 + x^2 = 40000\]
\[(3x+6)^2 + x^2 = 40000\]

Otherwise, you aren't squaring \(y\), you're squaring something different.

Answer 8

Next step is to expand that equation (multiply out \((3x+6)^2\))

Answer 9

9x + 36

Answer 10

No. \[(3x+6)^2 = (3x+6)(3x+6) = \]

Answer 11

oh lol ok

Answer 12

Multiply each part of the first thing by each part of the second thing, then collect like terms.

Answer 13

9x^2 + 18x + 18x + 18

Answer 14

edit: 36

Answer 15

9x^2 + 36x + 36

Answer 16

Even if the approach you used worked, it would be \(9x\color{red}{^2} + 36\), right? :-)

\[(3x+6)(3x+6) = 9x^2 + 18x + 18x + 36 = 9x^2 + 36x + 36\]

Answer 17

Okay, so now we have \[9x^2+36x + 36 + x^2 = 40000\]Can you solve that for \(x\)?

Answer 18

um

Answer 19

can i combine 9x^2 and x^2?

Answer 20

yes, you can and should do that.

Answer 21

so 10x^2 + 36x + 36= 40,000

Answer 22

what now ?

Answer 23

Yep. Do you know the quadratic formula?

if you have \[ax^2+bx+c =0, \, a \ne 0\]\[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 24

No two ways about it, this will be a bit ugly with that equation :-)

Answer 25

now sub ?

Answer 26

You have to rearrange your equation slightly to get it into that form. Do you see what to do?

Answer 27

Perhaps the first thing I would do is divide everything by 2:

\[5x^2 + 18x + 18 = 20000\]
Next, subtract 20000 from both sides so that you have the desired \(=0\) form:
\[5x^2+18x+18-20000 = 0\]\[5x^2+18x-19982=0\]\[a=5,b=18,c=-19982\]

Answer 28

ok …

Answer 29

then plug those numbers into that formula for the values of \(x\) where the spaceship's paths will intersect. you also need to find the corresponding values of \(y\) from your formula with those values of \(x\).

Answer 30

\[x= \frac{ -18\pm \sqrt{18^2 - 4(5)(-19982)} }{ 10 }\]

Answer 31

So far so good...now simplify that :-)

Answer 32

\[x = \frac{ -18\pm \sqrt{399,640} }{ 10}\]

Answer 33

the square root of 399,640 is 632.170862979

Answer 34

so \[x= \frac{ -18 \pm 632.170 }{ 10 }\]

Answer 35

right ? @whpalmer4

Answer 36

\[18^2-4(5)(-19982) = 324+399640 = 399964\]

Answer 37

so x = 399964

Answer 38

No. That's the quantity under the square root sign. You forgot the b^2 part in your calculation.

Answer 39

*facepalm*

Answer 40

I'll save you the trouble of trying to simplify the square root: 399964 = 4*99991 so the square root turns into \(2\sqrt{99991} \approx 632.427\) which is pretty close to what you had :-)

Answer 41

ok..

Answer 42

I've got to go now sorry ill be back and we can continue ok ? thanks!

Answer 43

ok back.

Answer 44

don't help this guy, he's reterderd

Answer 45

lol dude get a life pls

Answer 46

just trying to pass time bro x_x ill stop messing with you though

Answer 47

the equation is {(3x+4y=48,x-y=2) this is how the teacher wrote it out on my assignment. The answers given were x=40 and y=42 I need to understand how these numbers were gotten and the book is no help. I am visual and there are no examples close to this problem.
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Firstly, that's not an equation. It's 2 equations, called a system of equations.
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2nd, 40 & 42 are not the answers.
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3x+4y=48
x-y=2 --> x = y+2

Answer 48

Sub for x in the 1st eqn
3(y+2) + 4y = 48
6y+6 + 4y = 48
7y = 42
y = 6
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x = y+2
x = 8

Answer 49

what?

Answer 50

@whpalmer4 ?

Answer 51

x=8 P.P

Answer 52

i don't think we have the same assignment

Answer 53

same >.<

Answer 54

oh

Answer 55

Peter,

Since I haven't yet seen Parts 1 and 2 of this problem, I'm making the assumption that our astronaut approaches the circular orbit x^2 + Y^2 = 40000 along the path described by y = 3x + 6. I'm also assuming that Astro A. approaches from the left, as x increases. That means Astro's initial linear path intersects the circular orbit of radius 200 in Quadrant 2. With me so far?

Then all we have to do is to substitute y=3x+6 into x^2 + y^2 = 40000 = 200^2 and solve for x; one such value will be negative, the other positive (if my assumptions are correct). Take the negative one and substitute that for x in either y=3x+6 or x^2 + y^2 = 200^2. That'll give you a positive value for y.

The resulting point, (x,y), is the desired point of intersection.

How much sense does this make to you? What further info would you need, if any, or like to have, before you move on to Part 4?

MM