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In 2008, the Singapore Flyer opened up as the world's largest Ferris wheel. It has a maximum height of 165 meters and a diameter of 150 meters, with one full rotation taking about 30 minutes. Find an equation for the wheel if the center of the wheel is on the y-axis also known as the x = 0 line.
Let's see what I can do here.
If x = o, then the center is (0, y) = (h, k).
The height of the wheel is 165 meters which is the diameter.
I say the value of y is halfway up the wheel.
So, y = 165/2 = 82.5 meters.
This leads to the center point being (0, 82.5).
The radius is r.
r = (1/2)(150) = 75 meters.
I now have the radius and center point.
I think the next step is to plug this information into the standard form of an equation of a circle:
(x - h)^2 + (y - k)^2 = r^2
Is this correct so far?
Convert each equation in standard form to general form.
The general form of the equation of a circle is given sat?
x^2 + y^2 + ax + by + c = 0.
1. (x - 3)^2 + (y + 3)^2 = 9
2. (x + 1)^2 + (y - 2)^2 = 4
Question 1
(x - 3)^2 + (y + 3)^2 = 9
(x - 3)(x - 3) + (y + 3)(y + 3) = 9
x^2 - 6x + 9 + y^2 + 6y + 9 - 9 = 0
x^2 + y^2 - 6x + 6y + 18 - 9 = 0
x^2 + y^2 - 6x + 6y + 9 = 0
You say?
Question 2
(x + 1)^2 + (y - 2)^2 = 4
(x + 1)(x + 1) + (y - 2)(y - 2) = 4
x^2 + 2x + 1 + y^2 -4y + 4 - 4 = 0
x^2 + y^2 + 2x - 4y + 1 + 4 - 4 = 0
x^2 + y^2 + 2x - 4y + 5 - 4 = 0
x^2 + y^2 + 2x - 4y + 1 = 0
You say?
Find the standard form of the equation of the circle with endpoints of a diameter at (4, 3) and (0, 1).
I think the midpoint must be found here.
If that's the case, let M = midpoint.
M = (2, 2)
I now need r.
Let r = the distance from any of the endpoints of the circle to the midpoint.
I will use the endpoint (0, 1).
r = sqrt{(0 - 2)^2 + (1 - 2)^2}
r = sqrt{5}
Plug r into x^2 + y^2 = r^2.
x^2 + y^2 = (sqrt{5})^2
My answer is x^2 + y^2 = 5.
Yes?
amnkb wrote:All the factors are:
1, 2, 3, 4, 6, 12, 1x, 2x, 3x, 4x, 6x, 12x, 1x^2, 2x^2, 3x^2, 4x^2, 6x^2, 12x^2If I understood well the exercise, amnkb gave the right answer.
For example. let us assume x=5
12x^2 = 300
So, its factors are:
1, 2, 3, 4, 6, 12, 5, 10, 15, 20, 30, 60, 25, 50, 75, 100, 150, 300
----------------- ----------------------- -----------------------------
Ok. Sounds good. Thanks for your input.
Find the standard form of the equation of the circle with center (-3, 1) and tangent to the y-axis.
Let me see.
If the circle is tangent to the y-axis, then it must touch the line x = 0 at the point (0, 1).
I need to find r.
r = sqrt{(0 - (-3))^2 + (1 - 1)^2}
r = 3
x^2 + y^2 = r^2
x^2 + y^2 = (3)^2
The equation is x^2 + y^2 = 9.
You say?
Find the standard form of the equation of the circle centered at the origin and containing the point (-2, 3).
Let's do this one.
The standard form is x^2 + y^2 = r^2.
I need to find r.
r = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
r = sqrt{(-2 - 0)^2 + (3 - 0)^2}
r = sqrt{(-2)^2 + (3)^2}
r = sqrt{4 + 9}
r = sqrt{13}
Plug sqrt{13} into x^2 + y^2 = r^2 for r.
x^2 + y^2 = (sqrt{13})^2
x^2 + y^2 = 13
Yes?
FelizNYC wrote:TWO LOVES
1. MATHEMATICS
2. CLASSICAL GUITAR HYMNS
Hi, solo_guitar!
What? Who cares? Stick to question at hand.
FelizNYC wrote:Find the GCF of 45 and 75.
Another way is to do the prime factorization w/ a table:
factors 45: 3 * 3 * 5 75: 3 * 5 * 5 ---+-------------- GCF: 3 * 5 = 15
The prime factorization method is cool.
FelizNYC wrote:You know, the AI Questions app tells me that the factors are 1, 2, 3, 4, 6, x, 2x, 4x, 6x. I also know that my work is correct. Who's right?
you did the factoring right
But you didn't put things back together in *all* the possible ways
AI is many times stupid at math
But it is right-er this time
All the factors are:
1, 2, 3, 4, 6, 12, 1x, 2x, 3x, 4x, 6x, 12x, 1x^2, 2x^2, 3x^2, 4x^2, 6x^2, 12x^2
I checked with professor Leonard. He said my answer is correct.
x^2 - 2x + 1 + y^2 + 4y + 4 - 9 = 0
x^2 + y^2 - 2x + 4y + 1 + 5 - 9 = 0
The plus 4 on the correct line has become plus 5 on the next line.
Bob
Thank you. Someone else pointed out my typo. Rest assure that this is simply a self-study of math learned back in the 80s and 90s. What else is a middle-aged lonely guy to do?
TWO LOVES
1. MATHEMATICS
2. CLASSICAL GUITAR HYMNS
FelizNYC wrote:x^2 - 2x + 1 + y^2 + 4y + 4 - 9 = 0
x^2 + y^2 - 2x + 4y + 1 + 5 - 9 = 0x^2 - 2x + 1 + y^2 + 4y + 4 - 9 = 0
to
x^2 + y^2 - 2x + 4y + 1 + 5 - 9 = 0It is actually:
x^2 + y^2 - 2x + 4y + 1 + 4 - 9 = 0Your typo was 5 instead of 4.
I recall how, me too, I did this type of mistake in some math exams (when I was a student many decades ago). I even failed in one of them because of a silly typo I did.
I got it. Thanks. It was simply a typo.
FelizNYC wrote:(x - 1)^2 + (y + 2)^2 = 9
...
x^2 + y^2 - 2x + 4y - 3 = 0(x - 1)^2 + (y + 2)^2 = 9
Let us assume x=1:
(1 - 1)^2 + (y + 2)^2 = 9
(y + 2)^2 = 9
y + 2 = 3
y = 3 - 2
y = 1Now,
x^2 + y^2 - 2x + 4y - 3 = 0Let us verify if this final form is also satisfied for x=1 and y=1
1^2 + 1^2 - 2*1 + 4*1 - 3 = 0
1 + 1 - 2 + 4 - 3 = 1 ≠ 0I am afraid that you did a typo somewhere.
Can you show me where the error was made?
The standard form of an equation of the circle of radius r and center at the origin (0, 0) is x^2 + y^2 = r^2.
The general form of an equation of a circle is given by
x^2 + y^2 + ax + by + c = 0.
The standard form of an equation of a circle with radius r and center
(h, k) is given by (x - h)^2 + (y - k)^2 = r^2.
Write the standard form and the general form of the equation of each circle of radius r and center (h, k) .
r = 3; (h, k) = (1, -2)
Let me see.
This circle is not centered at the origin.
Plug r = 3, h = 1, and k = -2 into
(x - h)^2 + (y - k)^2 = r^2
(x - 1)^2 + (y - (-2))^2 = 3^2
(x - 1)^2 + (y + 2)^2 = 9....This is the standard form not centered at the origin.
Next, write the general form of the equation of the circle.
(x - 1)^2 + (y + 2)^2 = 9
(x - 1)(x - 1) + (y + 2)(y + 2) - 9 = 0
x^2 - 2x + 1 + y^2 + 4y + 4 - 9 = 0
x^2 + y^2 - 2x + 4y + 1 + 5 - 9 = 0
x^2 + y^2 - 2x + 4y - 3 = 0
You say?
FelizNYC wrote:In that case, the equation should be
x^2 + y^2 + 10x + 4y - 20 = 0
Yes?
Yes.
Perfect. I will post similar questions when time allows.
What does this mean? And does it have to be y?
By rewriting, for example, f(x)=2x+3, as y=2x+3, are we simply stating that something = 2x+3; and in the first case we’re calling that something f(x), and in the second case we’re calling it y?
Does the y have anything to do with the y axis in x,y coordinates? Or is just a randomly chosen letter? Could it just as well be z, or a, or b, etc?
Note: y and f(x) mean the same thing. It's like saying car and automobile.
For example, y = 2x + 5 can also be written as f(x) = 2x + 5. In terms of functions, f(x) is mostly used.
Q1 Perfect.
Q2. Almost. Correct up to this line:
x^2 + 10x + 25 + y^2 + 4y + 4 = 49
You've brought the 49 across to the left hand side but forgotten there's already +25 and + 4 there.
Bob
In that case, the equation should be
x^2 + y^2 + 10x + 4y - 20 = 0
Yes?
The general form of Q1 is:
x^2 + y^2 - 9 = 0
Cool. Thanks.
Maybe the problem here is that the word factor is being used to mean two similar but different things.
If N is a whole number then another whole number m is a factor if N/m is a whole number.
eg. 8 is a factor of 48 because 48/8 = 6
But if an algebraic expression (E) can be split into two expressions (F and G) that multiply to make E then F and G are said to be factors.
eg.
x and (x+2) are factors of E.
If only numbers are involved then the GCF is clear,
eg The GCF of 45 and 75 is 15
In pure mathematical terms saying that the GCF of 12x^2 and 28x^4 is 4x^2 is a questionable use of GCF because we don't even know if x is a whole number so how can we ask questions about the greatest factor.
But the Prof. goes on to use his earlier method to factorise (second definition) algebraic expressions. Nothing wrong with that; methods that work but have use dubious logic do exist in maths . I teach integration by saying have a guess based on what you know about differentiation; check if it works; and modify your guess if it doesn't.
Bob
The idea of GCF is basic. I don't think professor Leonard wants to go too deep mathematically in a course called Intermediate Algebra. Students in such a course are beginning algebra students.
The standard form of an equation of the circle of radius r and center at the origin (0, 0) is x^2 + y^2 = r^2.
The general form of an equation of a circle is given by
x^2 + y^2 + ax + by + c = 0.
The standard form of an equation of a circle with radius r and center
(h, k) is given by (x - h)^2 + (y - k)^2 = r^2.
Write the standard form and the general form of the equation of each circle of radius r and center (h, k) .
1. r = 3; (h, k) = (0, 0)
2. r = 7; (h, k) = (-5, -2)
Question 1
This circle is centered at the origin.
x^2 + y^2 = r^2
Let r = 3.
x^2 + y^2 = (3)^2
x^2 + y^2 = 9===>standard form
How do I write x^2 + y^2 = 9 in the general form
x^2 + y^2 + ax + by + c = 0?
Question 2
r = 7; (h, k) = (-5, -2)
This circle is not centered at the origin.
(x - h)^2 + (y - k)^2 = r^2
(x - (-5))^2 + (y - (-2))^2 = (7)^2
(x + 5)^2 + (y + 2))^2 = 49
(x + 5)(x + 5) + (y + 2)(y + 2) = 49
x^2 + 10x + 25 + y^2 + 4y + 4 = 49
I need to write it in the form x^2 + y^2 + ax + by + c = 0.
My answer is: x^2 + y^2 + 10x + 4y - 49 = 0.
You say?
Watched it. Great vid!
But I didn't hear him say what he means by 'evenly divided'. It's not a maths expression that I'm familiar with.
Did it come from a text book question?
Bob
Evenly divided are my words to mean there is no remainder.
For example: 3x^2 ÷ x = 3x because 3x times x = 3x^2.
Given 3x^2 and 3x, find the GCF.
The GCF IS 3x. Why?
I say 3x is the GCF because 3x^2 can be evenly divided (no remainder) by 3x.
We can divide 3x^2 by 3x but the other way around.
OK, thanks for the link. I really like this guy's teaching style; it must be great to be in his class.
He is trying to find common factors so it is sufficient to find x, x^2, x^3 etc and not bother about 3x, 3x^2 and 3x^3.
He takes care of the number factors separately so it works.
So it all depends on what the question asks for.
If it just wants the factors then 3x etc should be in the list, but if you're after the GCF you can leave them out.
Bob
I agree. Professor Leonard is an amazing teacher. He wants students to understand math and not just to know enough to pass the test or the class.
Relatively prime means the two (or more) numbers have only 1 as a common factor.
For example: 6(= 2 x 3) and 11 are relatively prime, 45( = 3 x 3 x 5) and 77( = 7 x 11) are relatively prime
6 and 15 are not; as 3 is also a factor of both.
If you factorise out the GCF from an expression then the numbers left must have no common factor because if there was another factor for both then you haven't found the greatest common factor.
So the numbers left are relatively prime.
I'll make up another example.
42x + 70 = 2(21x + 35) 21 and 35 are not relatively prime, but I haven't yet factored out all the common factors.
Let's do that:
42x + 70 = 2(21x + 35) = 14(3x + 5) Now we have 3 and 5, and these are relatively prime.
Bob
Thanks. Easy to understand now.
Factor 49x - 35.
This is what professor Leonard did:
Divide each term by the GCF.
The GCF is 7.
49x ÷ 7 = 7x
-35 ÷ 7 = -5
7(7x - 5)....done!
However, he went on to say that the binomial inside the parentheses is called RELATIVELY PRIME. Leonard stated that the binomial inside the parentheses will always be relatively prime.
What does he mean by those words?
Correct.
Bob
Like Richie Cunningham is famous for saying the line on Happy Days:
"I'm cooking with gas."
Yes for both.
Bob
It certainly feels good to know that I am right this time around. Math helps keep my brain cells alive. No pun intended.