The Official Homework Help Thread
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Superpenguin
Penguins! <3
If it's dealing with Mechanics, I can help you. I'm not learning E/M until this year.
Awesome! Yeah, I'm in mechanics and will do E/M during second semester. So far it's review, nothing difficult just yet.
ok hi so im doing a summer packet for school and i really have like no idea how to do these types of questions so if someone could like explain it that would be amazing!!!
the questions are "write an equation in slope intercept form for the line that contains the given point and is parallel to the given line" (ex. (1,4); y=-3x+2)
and "write an equation in slope intercept form for the line that contains the given point and is perpendicular to the given line" (ex. (2,4); y=1/2x+3)
sorry if that's a lot but i totally forgot how to do all of that over the past two years and would love some help c:
the questions are "write an equation in slope intercept form for the line that contains the given point and is parallel to the given line" (ex. (1,4); y=-3x+2)
and "write an equation in slope intercept form for the line that contains the given point and is perpendicular to the given line" (ex. (2,4); y=1/2x+3)
sorry if that's a lot but i totally forgot how to do all of that over the past two years and would love some help c:
aericell
☆ happinessdelight ☆
hi! perpendicular lines have a negative & reciprocal slope while parallel lines have the same. so all you need to do is plug the point into y = mx + b with the slopes (m) and solve for b to complete the equation
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A line in slope-intercept form can be written as y = mx + b where m = the slope and b = the y-coordinate of the y-intercept. Two lines are said to be parallel if their slopes are the same, and two lines are said to be perpendicular if their slopes are negative reciprocals of one another (e.g, if one line has slope 2 then a line that is perpendicular must have a slope of -1/2).
In both of your questions, you are aiming to find slope-intercept form equations of lines that satisfy the conditions of (1) passing through a specific point, and (2) that are either parallel or perpendicular to another line. Hence, you want to solve y = mx + b for m and b that meet these conditions.
Here's an example question that is similar to your first. Suppose that you want to find the equation of a line that passes through the point (1, 2) and is parallel to the line y = 3x + 1. Since this line must be parallel to y = 3x + 1 we have that the slope of this mystery line is m = 3. So plugging this into y = mx + b gives us:
y = 3x + b
Notice that any and all lines of the form above will be parallel to y = 3x + 1 since their slopes are the same. All we need to do now is solve for b. We're given that the line must pass through the point (1, 2) and so by substituting x = 1 and y = 2 into the equation above, we can solve for b:
(2) = 3(1) + b
2 = 3 + b
-1 = b
Substituting b = -1 into the original equation gives us the equation of the line that passes through the point (1, 2) and is parallel to y = 3x + 1:
y = 3x - 1
I hope that this helps! You should be able to complete both of the questions you have above if you follow a similar manner to this example. Just be weary that when you're finding the equation of a line that passes through a point and is perpendicular (instead of parallel) to another line, then the slope m of your answer will be the negative reciprocal of the slope of the line in your problem.
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I was basically just digging up my old french textbook and excercises book since I want to keep my French skills on the same level now I dropped french this year since I don't feel like I'm advanced enough to do exams in it. But I'm a bit confused about what to do with this sentence: "Je (ecouter) de la musique quand quelqu'un (frapper) a la porte." I'm having a dilemma on if I need to use the Pass? Compos? on the first verb and the Imparfait on the second or the opposite? I'm pretty sure I don't need to use the same tense for both, right? At least not in this case, I thought.
Also, what do I need to do with the verbs pouvait and avait in the imparfait? I'm pretty sure this isn't something I've had before. ''Mais Marie a r?pondu qu'elle ne (pouvait) pas aller parce qu'il (avait) un rendez-vous'' Do I've to follow special rules because the verbs are related to each other? Uhg, maybe it's for the better that I dropped it.
Also, what do I need to do with the verbs pouvait and avait in the imparfait? I'm pretty sure this isn't something I've had before. ''Mais Marie a r?pondu qu'elle ne (pouvait) pas aller parce qu'il (avait) un rendez-vous'' Do I've to follow special rules because the verbs are related to each other? Uhg, maybe it's for the better that I dropped it.
Ok here's a quick physics problem.
It states that the answer to 10 is B, but that doesn't really make sense for me. Shouldn't the electric field magnitude be zero at x = 0 because the electric field strength +Q would be kq/r^2 and from -Q would be -kq/r^2? If you combine those two magnitudes it would come out to be zero. I got D for this because when a positive charge were to go from -a to a, then there would be two vectors going to the right (because the top and bottom vectors cancel out) until the charge goes to x = 0 where the total magnitude would be zero. Is my thinking correct or am I missing something here...
Also the same thing for 12. It asks for the electric force when a negative charge goes from -a to a, shouldn't the forces cancel out at 0?
It states that the answer to 10 is B, but that doesn't really make sense for me. Shouldn't the electric field magnitude be zero at x = 0 because the electric field strength +Q would be kq/r^2 and from -Q would be -kq/r^2? If you combine those two magnitudes it would come out to be zero. I got D for this because when a positive charge were to go from -a to a, then there would be two vectors going to the right (because the top and bottom vectors cancel out) until the charge goes to x = 0 where the total magnitude would be zero. Is my thinking correct or am I missing something here...
Also the same thing for 12. It asks for the electric force when a negative charge goes from -a to a, shouldn't the forces cancel out at 0?
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aericell
☆ happinessdelight ☆
Hi! Is anyone good at Calculus? My brother asked me for help on a problem but I don't remember learning how to do this. My teacher last year only prepared us for the AP test so I think he might have skipped a few things.
Integrate 1/(x^2 + 9)
If someone could please explain how to integrate this or something similar to it please! Tyvm~
Integrate 1/(x^2 + 9)
If someone could please explain how to integrate this or something similar to it please!
Tyvm~
Hi! Is anyone good at Calculus? My brother asked me for help on a problem but I don't remember learning how to do this. My teacher last year only prepared us for the AP test so I think he might have skipped a few things.
Integrate 1/(x^2 + 9)
If someone could please explain how to integrate this or something similar to it please!
So for a function in the format 1/(x^2 + a^2) the integral with respect to x is 1/a arctan(x/a). So in this case a=3 (and a^2 = 9) so it would integrate as: [arctan(x/3)]/3 +C.
Hope that was right/made sense my calculus is a little rusty
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Hi! Is anyone good at Calculus? My brother asked me for help on a problem but I don't remember learning how to do this. My teacher last year only prepared us for the AP test so I think he might have skipped a few things.
Integrate 1/(x^2 + 9)
If someone could please explain how to integrate this or something similar to it please!
I will answer the similar problem of integrating f(x) = 1/(x^2 + 4) and your brother should be able to integrate his homework problem by following the same steps ^^.
It's really important to notice that this function can't really be integrated by any of the typical integration rules. Furthermore, the denominator is of the form "x^2 + a^2" where in this case a = 2. When you have to integrate functions involving forms like "x^2 + a^2", "x^2 - a^2", or "a^2 - x^2" it is important to immediately recognize to use a trigonometric substitution. In general:
If you see x^2 + a^2 then use the substitution x = atanθ for -π/2 ≤ θ ≤ π/2.
If you see x^2 - a^2 then use the substitution x = asecθ for 0 ≤ θ ≤ π/2.
If you see a^2 - x^2 then use the substitution x = asinθ for -π/2 ≤ θ ≤ π/2.
Here we have the first form, so we will immediately let x = 2tanθ (noting that a=2 for this example). Then dx/dθ = 2sec^2 θ, so dx = 2sec^2 θ dθ and applying this substitution gives:
Now use the trigonometric identity for which tan^2 θ + 1 = sec^2 θ to get:
Now since we let x = 2tan θ we have that arctan(x/2) = θ. So:
I hope this help! Don't forget to tack on the +C arbitrary constant on at the end!
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aericell
☆ happinessdelight ☆
so is it just something you kinda have to have memorized? will the answer be the same format for these types of problems?
but thank you both so much for your help! hopefully he understands it ^^
