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[Bobkitty246]

TYPE OF QUESTION:math
YOUR QUESTION:

Either I am not understanding this question the way it was supposed to be understood, and made things more difficult for myself, or I am just stupid ouo. I'd like to know your understandings for this question hehe! ^^" (its the first question on the page with the image next to it) sorry for the countless questions!

[wane]

TYPE OF QUESTION: math
YOUR QUESTION:
so there's a function and I'm supposed to find the vertex of the parabola without graphing it.
the function is: f(x)=-(1/2x+1)^2+1
I know the vertex is the horz. shift and the vert. shift, but I'm unsure why the vertex is at (-2,1) and not (1,1)(using the horz. shift and vert. shift)
Could someone explain it to me really throughly how to solve it and why? Sorry if this seems really easy, I'm bad at math and I have a test on this tomorrow. e-e Thanks for any help!

[M00N]

TYPE OF QUESTION:math
YOUR QUESTION:

Either I am not understanding this question the way it was supposed to be understood, and made things more difficult for myself, or I am just stupid ouo. I'd like to know your understandings for this question hehe! ^^" (its the first question on the page with the image next to it) sorry for the countless questions!

I think you're dealing with the Pythagorean theorem, no?
(a²+b²=c²)
in this case dougie is known to be in the center of the posts, and jon is at one end of an iso triangle, so the midline (jon to dougie) that divides the triangle forms two right triangles
now you plug in the lengths into the equation
it's been so long since doing this for me all I can think of is cos/tan/sine though?
It may be like
(3x²+x²=2.8²)
(4x²=2.8²) ect

old:

I think you're dealing with the Pythagorean theorem, no?
(a²+b²=c²)
in this case dougie is known to be in the center of the posts, and jon is at one end of an iso triangle, so the midline (jon to dougie) that divides the triangle forms two right triangles
now you plug in the lengths into the equation
(3x² + x² = 2.8²)
(4x²=7.84)
divide both sides by 4
(x²=1.96)
take the square root of each side
(x=1.4)
just one more step! since the net is 2x wide, its width= 2 times x = 1.4 times 2 = 2.8 m
and theres your answer and how to get it!

[Rabid_Jaguar]

TYPE OF QUESTION: math
YOUR QUESTION:
so there's a function and I'm supposed to find the vertex of the parabola without graphing it.
the function is: f(x)=-(1/2x+1)^2+1
I know the vertex is the horz. shift and the vert. shift, but I'm unsure why the vertex is at (-2,1) and not (1,1)(using the horz. shift and vert. shift)
Could someone explain it to me really throughly how to solve it and why? Sorry if this seems really easy, I'm bad at math and I have a test on this tomorrow. e-e Thanks for any help!

I'm unclear where it says that the vertex is (-2,1). Seeing this problem, this is what I think:

So the vertex form is y = a(x - h)^2 +k

The vertex of a function comes from the variables (-h,k). In your equation, h = 1 and k = 1. Therefore, your vertex would be (-1,1), since you take the opposite of the h term.

Bobkitty, I'm so sorry but I only ever did a couple of those types of problems, and that was several years ago. I'm not honestly sure how to solve that problem either. I would say to ask your friends about it, and if they can't help, ask your teacher. If the teacher is any good, he or she take the time to help you if you don't understand something.

EDIT: I think M00N is correct, the math looks right and that assumption can be made with the information given.

[kaotically]

TYPE OF QUESTION:math
YOUR QUESTION:

Either I am not understanding this question the way it was supposed to be understood, and made things more difficult for myself, or I am just stupid ouo. I'd like to know your understandings for this question hehe! ^^" (its the first question on the page with the image next to it) sorry for the countless questions!

I think you're dealing with the Pythagorean theorem, no?
(a²+b²=c²)
in this case dougie is known to be in the center of the posts, and jon is at one end of an iso triangle, so the midline (jon to dougie) that divides the triangle forms two right triangles
now you plug in the lengths into the equation

But your implying that all the sides are the same length. It would be a perfect triangle if that was true. (Please correct me if I'm wrong in thinking this)

[Faisre]

TYPE OF QUESTION:math
YOUR QUESTION:

Either I am not understanding this question the way it was supposed to be understood, and made things more difficult for myself, or I am just stupid ouo. I'd like to know your understandings for this question hehe! ^^" (its the first question on the page with the image next to it) sorry for the countless questions!

I think you're dealing with the Pythagorean theorem, no?
(a²+b²=c²)
in this case dougie is known to be in the center of the posts, and jon is at one end of an iso triangle, so the midline (jon to dougie) that divides the triangle forms two right triangles
now you plug in the lengths into the equation
(3x² + x² = 2.8²)
(4x²=7.84)
divide both sides by 4
(x²=1.96)
take the square root of each side
(x=1.4)
just one more step! since the net is 2x wide, its width= 2 times x = 1.4 times 2 = 2.8 m
and theres your answer and how to get it!

Hope you don't mind me intruding on this thread. I've been feeling like I'm getting dumber lately, so I thought I'd peek in and see what sort of things were being asked here.

Anyway, I'm not so sure about using the Pythagorean Theorem here. Now, admittedly, I haven't done math like this in at least 7 years (shhh, I'm not that old, I promise), but based on the illustration and the word problem, I'm not sure that Dougie IS in the center of the posts. He is in front of the goal in the image, therefore making an obtuse triangle, not a right triangle. The word problem only says that Johnny is 3 times as far from the posts as Dougie, but not that he is necessarily directly between the poles? Maybe I'm just making it more complicated than it should be, though.

[M00N]

But your implying that all the sides are the same length. It would be a perfect triangle if that was true. (Please correct me if I'm wrong in thinking this)

This is true, but I think that it can still be equilateral like that?

aa, I wish I would have saved my geometry notes from the course. it had the specifics of this

TYPE OF QUESTION:math
YOUR QUESTION:

Either I am not understanding this question the way it was supposed to be understood, and made things more difficult for myself, or I am just stupid ouo. I'd like to know your understandings for this question hehe! ^^" (its the first question on the page with the image next to it) sorry for the countless questions!

I think you're dealing with the Pythagorean theorem, no?
(a²+b²=c²)
in this case dougie is known to be in the center of the posts, and jon is at one end of an iso triangle, so the midline (jon to dougie) that divides the triangle forms two right triangles
now you plug in the lengths into the equation
(3x² + x² = 2.8²)
(4x²=7.84)
divide both sides by 4
(x²=1.96)
take the square root of each side
(x=1.4)
just one more step! since the net is 2x wide, its width= 2 times x = 1.4 times 2 = 2.8 m
and theres your answer and how to get it!

Hope you don't mind me intruding on this thread. I've been feeling like I'm getting dumber lately, so I thought I'd peek in and see what sort of things were being asked here.

Anyway, I'm not so sure about using the Pythagorean Theorem here. Now, admittedly, I haven't done math like this in at least 7 years (shhh, I'm not that old, I promise), but based on the illustration and the word problem, I'm not sure that Dougie IS in the center of the posts. He is in front of the goal in the image, therefore making an obtuse triangle, not a right triangle. The word problem only says that Johnny is 3 times as far from the posts as Dougie, but not that he is necessarily directly between the poles? Maybe I'm just making it more complicated than it should be, though.

In the question itself, it describes doug as being "(whatever) from either post", implying that he is the center of the posts
I get what youre saying, but I think this may fall under section 3 of the question. without him being in a straight line between the posts I dont think its solvable without angle measurements

[Rabid_Jaguar]

To add on to what M00N said, the question itself asks "Discuss any assumptions you had to make." Like Faisre, it's been a while since I've done this kind of math, but I'm not sure there IS another way to solve this without making some small assumptions - I don't think we're given enough information otherwise.

And Shadowblaze, the Pythagorean Theorem doesn't imply that all the sides are the same length, just two of them (which is one of the things the problem tells us).

[Faisre]

In the question itself, it describes doug as being "(whatever) from either post", implying that he is the center of the posts
I get what youre saying, but I think this may fall under section 3 of the question. without him being in a straight line between the posts I dont think its solvable without angle measurements

To add on to what M00N said, the question itself asks "Discuss any assumptions you had to make." Like Faisre, it's been a while since I've done this kind of math, but I'm not sure there IS another way to solve this without making some small assumptions - I don't think we're given enough information otherwise.

And Shadowblaze, the Pythagorean Theorem doesn't imply that all the sides are the same length, just two of them (which is one of the things the problem tells us).

Good point. Unless I'm forgetting something, without an angle you can't use COS or any of those things, which would be the way to solve an obtuse triangle if I'm not mistaken?

In that case, if we do go with the right angle assumption, I believe I agree with MOON's math.

[Rabid_Jaguar]

In the question itself, it describes doug as being "(whatever) from either post", implying that he is the center of the posts
I get what youre saying, but I think this may fall under section 3 of the question. without him being in a straight line between the posts I dont think its solvable without angle measurements

To add on to what M00N said, the question itself asks "Discuss any assumptions you had to make." Like Faisre, it's been a while since I've done this kind of math, but I'm not sure there IS another way to solve this without making some small assumptions - I don't think we're given enough information otherwise.

And Shadowblaze, the Pythagorean Theorem doesn't imply that all the sides are the same length, just two of them (which is one of the things the problem tells us).

Good point. Unless I'm forgetting something, without an angle you can't use COS or any of those things, which would be the way to solve an obtuse triangle if I'm not mistaken?

In that case, if we do go with the right angle assumption, I believe I agree with MOON's math.

Hopefully, Bobkitty, all this mess of "this is right" "that is right" helped you out a bit. c: It's no problem having lots of math questions - you should have seen all the questions I was asking not too long ago. cx