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Random Math Questions

M

manlymatt83

Lifer
I'm helping my friend with a big math question booklet later today and I had a few questions about some of the problems, reviewing my own "knowledge".

1) First, to find the equation of the line that passes through the point (0,4) and is perpendicular to the line 5x + 2y = 3, you just solve for that equation normally, then flip the slope, correct?


2)

2 2 10
------- + ------- = ----------------
t - 3 t - 2 t^2 - 5t + 6


My assumption here is to simplify the polynomial on the right, then multiply all three by the output of that, and we can most likely cancel the bottom on the left hand side. Is that correct?



3)

SQR(x + 10) = x - 2

Here, I squared both sides, leaving me with:

x + 10 = (x - 2)(x - 2)

which I simplified to

x + 10 = x^2 - 4x + 4

Not sure where to take it from there though....

The others seem OK. Any help would be appreciated with confirming my guesses.
 
2) The right side of the equation is a quadratic equation. That should help solve that side, or at least where to go with it from there.
 
3) x + 10 = x^2 - 4x + 4

0 = x^2 - 5x -6

(-b +- sqrt(b^2 - 4ac)) / 2a.

x = __
x = __


... or you can just factor it. It factors real easy. 🙂

Edit: NVM, doesn't factor, I don't think.
Edit2: DOH. Yes it factors extremely easily.

1)

(Correct me if I'm wrong. It's been a hella long time since I've done this.)

5x + 2y = 3

y = -(5/2)x + (3/2)

m of the line perpendicular = (2/5)

y = (2/5)x + b

Substitute in points: 4 = (2/5)(0) + b

b = 4

y = (2/5)x + 4
 
its just 6 and -1, so factor is easiest. Although, personally I am a big time T&E man, so if you dont know an answer just blug numbers in till you get it, that took me 5 seconds to get the answer without having to do any math.
 
OK, and am I right with #1?

Also, one more question thatr I encountered. EVERYTHING else is OK.

Cube Root of 4x^2y multipled by Cube Root of 2x ^ 5 y ^ 2

Since Cube roots cancel out, you're left with:

4x^2 y * 2x^5 y^2

Which would be:

6x ^ 7 * y ^ 3

Correct? 🙂

I think that just about does it.
 
Originally posted by: mjuszczak
I'm helping my friend with a big math question booklet later today and I had a few questions about some of the problems, reviewing my own "knowledge".

1) First, to find the equation of the line that passes through the point (0,4) and is perpendicular to the line 5x + 2y = 3, you just solve for that equation normally, then flip the slope, correct?

Equation of this new line will be:
-2x + 5y = c. Substitute the point (0,4) to find c.
c = 20
=> -2x + 5y = 20 or alternatively 2x -5y = -20

Edit: Remember, if line (1) has slope m and line (2) is perpendicular to line (1), then line (2) will have slope -1/m, i.e. the negative reciprocal.



2)

2 2 10
------- + ------- = ----------------
t - 3 t - 2 t^2 - 5t + 6


My assumption here is to simplify the polynomial on the right, then multiply all three by the output of that, and we can most likely cancel the bottom on the left hand side. Is that correct?

Factor the denominator on the right side

2 2 10
------- + ------- = ----------------
t - 3 t - 2 (t-2)(t-3)

Multiply both sides by (t-2)(t-3)

-> 2(t-2) + 2(t-3) = 10 -> Solve for t.

3)

SQR(x + 10) = x - 2

Here, I squared both sides, leaving me with:

x + 10 = (x - 2)(x - 2)

which I simplified to

x + 10 = x^2 - 4x + 4

Not sure where to take it from there though....

Bring everying to one side:
x^2 -5x -6 =0
Factor:
(x-6)(x+1) = 0
x = 6 or x = -1

The others seem OK. Any help would be appreciated with confirming my guesses.
Click to expand...

 
Originally posted by: hypn0tik
Originally posted by: mjuszczak
I'm helping my friend with a big math question booklet later today and I had a few questions about some of the problems, reviewing my own "knowledge".

1) First, to find the equation of the line that passes through the point (0,4) and is perpendicular to the line 5x + 2y = 3, you just solve for that equation normally, then flip the slope, correct?

Equation of this new line will be:
-2x + 5y = c. Substitute the point (0,4) to find c.
c = 20
=> -2x + 5y = 20 or alternatively 2x -5y = -20

Edit: Remember, if line (1) has slope m and line (2) is perpendicular to line (1), then line (2) will have slope -1/m, i.e. the negative reciprocal.



2)

2 2 10
------- + ------- = ----------------
t - 3 t - 2 t^2 - 5t + 6


My assumption here is to simplify the polynomial on the right, then multiply all three by the output of that, and we can most likely cancel the bottom on the left hand side. Is that correct?

2 2 10
------- + ------- = ----------------
t - 3 t - 2 (t-2)(t-3)

Multiply both sides by (t-2)(t-3)

-> 2(t-2) + 2(t-3) = 10 -> Solve for t.

3)

SQR(x + 10) = x - 2

Here, I squared both sides, leaving me with:

x + 10 = (x - 2)(x - 2)

which I simplified to

x + 10 = x^2 - 4x + 4

Not sure where to take it from there though....

Bring everying to one side:
x^2 -5x -6 =0
Factor:
(x-6)(x+1) = 0
x = 6 or x = -1

The others seem OK. Any help would be appreciated with confirming my guesses.
Click to expand...
Click to expand...


Thanks!

For the first one, couldn't I just solve first then flip the slope and sign?

5x + 2y = 3
2y = -5x + 3
y = -5/2x + 1.5

m = -5/2

Therefore the equation of the line perpendicular would be:

y = 2/5x + 1.5
 
Originally posted by: mjuszczak

...Snip...
Click to expand...


Thanks!

For the first one, couldn't I just solve first then flip the slope and sign?

5x + 2y = 3
2y = -5x + 3
y = -5/2x + 1.5

m = -5/2

Therefore the equation of the line perpendicular would be:

y = 2/5x + 1.5[/quote]

Be careful. For the new line, you ONLY know the slope and a point that the line passes through. You cannot automatically say that the intercept is 1.5. You have to solve for the intercept.

i.e. y = 2/5x + b
Substitute (0,4)
b = 4

So the equation of the line perpendicular to the original and passing through (0,4) is y = 2/5x + 4
 
Originally posted by: hypn0tik
Originally posted by: mjuszczak

...Snip...
Click to expand...


Thanks!

For the first one, couldn't I just solve first then flip the slope and sign?

5x + 2y = 3
2y = -5x + 3
y = -5/2x + 1.5

m = -5/2

Therefore the equation of the line perpendicular would be:

y = 2/5x + 1.5
Click to expand...

Be careful. For the new line, you ONLY know the slope and a point that the line passes through. You cannot automatically say that the intercept is 1.5. You have to solve for the intercept.

i.e. y = 2/5x + b
Substitute (0,4)
b = 4

So the equation of the line perpendicular to the original and passing through (0,4) is y = 2/5x + 4[/quote]


Got ya, thanks
 
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