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Don't get into an argument with an idiot.

They drag you down to their level, then beat you on experience.

I have a few questions on that answer.

When I square \(a^2+b^2+c^2+1-d\), I get \(a^4+b^4+c^4+a^2b^2+a^2c^2+2a^2-2a^2d+b^2a^2+b^2c^2+2b^2-2b^2d+c^2a^2+c^2b^2+2c^2-2c^2d+d^2-2d+1\).

How did you get \(a^2+b^2+c^2+1−2d+d^2\)?

Where did you get \(a^2+b^2+c^2=1\)?

Where did you get \(a+b+c-d=0\)?

And if those are both true, then wouldn't d be 2?

The shortest possible distance is 5 + 4 + 5 = 14.

The equation a^2+b^2+c^2+1=d+sqrt(a+b+c−d)​ can be rewritten as a^2+b^2+c^2+1−d=sqrt(a+b+c−d)​. Squaring both sides, we get a^2+b^2+c^2+1−2d+d^2=a+b+c−d. Simplifying, we get a^2+b^2+c^2+d^2−2d=a+b+c.

Now, we can use the fact that a^2+b^2+c^2=1 to substitute for a^2+b^2+c^2 in the equation above. This gives us 1+d^2−2d=a+b+c.

Finally, we can use the fact that a+b+c−d=0 to substitute for a+b+c in the equation above. This gives us 1+d^2−2d=0.

Solving for d, we find that d = (1+/- sqrt(5))/2.

However, we know that d must be a real number, so the only possible value of d is (1+sqrt(5))/2​​.

Therefore, the value of d is (1+sqrt(5))/2, which is the golden ratio.  Nice!

Say the common ratio is r. Then \(a_1-4r=a_5, \space a_1-2r=a_3, \space a_1+a_3+a_5=3a_1-6r\)and \(a_1a_3a_5=a_1(a_1-2r)(a_1-4r)\)

This is a system of equations. The first one reveals \(a_1=2r-4\). The second one then becomes

\((2r-4)\cdot-4\cdot(-4-2r)=80\)

\(2\cdot-4\cdot-2\cdot(r-2)(r+2)=80\)

\((r-2)(r+2)=5\)

\(r=3\)

So \(a_1=2, \space a_2=5, \space a_3=8, \space a_5=14, \space a_{10}=29\)

So the answer you are looking for is 29.

Charlie and Riordan initially had different numbers of stickers even though they ended up with the same number of stickers after Seth gave them each a portion of his stickers. This is because the fractions of Seth's stickers that he gave to Charlie and Riordan were different.

After answering the question, we can state that probability in a circle = (angle where B can exist)/(Total Angle) => P = 2/3

Explanation:

What is a circle?

A circle is formed by every point in the plane that is a certain distance away from another point (center). Thus, it is a curve formed by points moving in the plane at a constant distance from a fixed point. It is also rotationallysymmetric about the center at all angles. A circle is a two-dimensional closed object in which every pair of points in the plane is equally spaced from the "center." A specular symmetry line is formed by a line that goes through the circle. It is also rotationally symmetric about the center at all angles.

We now want A and B to be separated by a distance greater than r. As a result, the angle should be greater than 60 degrees because the radius is equal to 60 degrees and decreases as the angle increases. As a result, point B cannot be in the 60-degree zone of A on both sides, that is, to A's left and right. 

As a result, probability = (angle where B can exist)/ (Total Angle)

P = (360 - (60)-(60))/360

P = 2/3

28

Explanation:

In order to solve this problem we need to find out what $QR$ is.

We know that the perimeter of the parallelogram was given as $24$ and that two of the sides were given as $5$. $5+5=10$ $24-10=14/2=7$. Now we know that the base was $7$.

We can find the height by taking the area that they gave us and dividing that by $7$. We get $4$.

Since we now know that the height of the parallelogram is $4$ that means $OQ$ is also $4$.

Next we need to find what $RT$ is.  

$4^2+B^2=5^2$.

Since we now $5^2$ is $25$ and that $4^2$ is $16$, we can use the following equation.

$25-16=9$. $\sqrt{9}=3$.  

Now we know that $RT$ is $3$ we can start adding the sides.

$7+3=10+4=14 \cdot 2=28$.

[b]The perimeter of the rectangle $QOST$ is 28![/b]

The product of the fractions is 7/2​. The numerators and denominators of the fractions form an arithmetic sequence with a common difference of 1. The first numerator is 2 greater than the first denominator. Therefore, the first numerator is 3 and the first denominator is 1. The answer is 3/1.

When written out in a string, your expression is:

 \(3(2n–4)– 1 4 n= 3 4 (n+4) n\)

\(6n-12-14n=34n^2+136n\)

\(34n^2+144n+12=0\)

\(17n^2+72n+6=0\)

\(n = {-72 \pm \sqrt{72^2-4\cdot17\cdot6} \over 34} \space \text{(Using the quadratic formula)}\)

\(n=\frac{-72 \pm \sqrt{4776}}{34}\)

\(n=\frac{-72 \pm 2\sqrt{1194}}{34}\)

\(n=\frac{-36 \pm \sqrt{1194}}{17}\)

Which is as simple as we can get it.

We may make use of the rule that if N's digits add up to a multiple of 3, N is a multiple of 3.

Since 6+7+8+9=30, which is a multiple of 3, all such N will be multiples of 3. Therefore the number you are looking for is the ways to organize the digits in N, which is 4!=24.

\(\displaystyle \frac{2x^{2}+2x-2}{x(x^{2}-1)}\equiv \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\)

Begin by putting the terms on the rhs over a common denominator.

\(\displaystyle \frac{2x^{2}+2x-2}{x(x^{2}-1)} \equiv \frac{A(x-1)(x+1)+Bx(x+1)+Cx(x-1)}{x(x^{2}-1)}\)

For the rhs to be identical to the lhs we need

\(\displaystyle 2x^{2}+2x-2 \equiv A(x-1)(x+1)+Bx(x+1)+Cx(x-1),\)

(the bottom lines are already identical).

There are two ways of proceeding, either collect up terms on the rhs and equate coefficients, 

\(\displaystyle 2x^{2}+2x-2 \equiv x^{2}(A+B+C)+x(B-C)+(-A) \\ 2 = A+B+C \\2 = B-C \\ -2 = -A \\ A=2, \qquad B = 1, \qquad C=-1,\)

or, quicker, substitute convenient values for x,

\(\displaystyle x=0: -2=-A , \qquad A=2,\\ x=1 : \;\;\;2=2B, \qquad \;\;\;B=1 \\ x=-1: -2=2C, \qquad C=-1.\)

Thanks Juriemagic but getting stuck on the wrong method happens to me quite often too.

Sometimes I can see the answer or at least an approach fairly quickly but that is just from years of exposure and practice.

Also teaching  helps you cement a deeper understanding. Sort of knowing it and muffling your way through may be good enough on a personal level but if you are teaching it effectively you really have to understand it at a deeper, more organised level.  Teaching a topic always made me think about it more.

Nice work Yesoka,

I got the same answer via a different method.

Use pythagoras theorem twice to set up 2 equations

Sove simultaneously first to get x and then to get the height

Good morning Melody,

I trust you are well and that things are going great for you.....Thank you for the explanation....You know, I knew there was a reason for that and I kind of understood that the cube function was not symmetrical....I explained it to my student my way, but realised I needed to get a more mathematical approach to explain this, which you did, and I am grateful....Thank you so very much Melody....

You know, I have come to realise that with you and all the other moderators on here, whenever you see a problem, or get some kind of question, the computer kicks in, almost like AI, and basicslly all kinds of approaches are computed, including limitations, with reasons, and a final answer or explanation surfaces within moments.

As for me....well that simply does not happen.......I may think of one thing only...and NOT get the result I want, or the explanation to a question I want....LOL I believe you understand what I'm babbling on about here......

I really take my hat of to you and all the other great people on this forum!!..Thank you  

0.25 

Grogg's dad is 55.

The answer is

[4 0 1]

[8 0 -2]

[12 0 8]

It is not there.

The answer is 42.

(a) No, 1/4 is not in the set S.

We can write 1/4 as follows:

1/4=1/3​+1/3​

Since 1 and -1 are the only possible values for ai​, we can see that the only way to represent 1/4 as a sum of the form a1/3​​+a2/3^2​​+a3/3^3​​+... is to have a1​=a2​=1. However, this is not possible, since ai​ can only be 1 or -1. Therefore, 1/4 is not in the set S.

(b) Yes, 1/7 is in the set S.

We can write 1/7 as follows:

1/7 = 1/3 - 2/3^2

Since 1 and -1 are the only possible values for ai​, we can see that the only way to represent 1/7 as a sum of the form a1/3​​+a2​/3^2​+a33^2​​+... is to have a1​=1 and a2​=−2. This is possible, since ai​ can be 1 or -1. Therefore, 1/7 is in the set S.

a=productfor(n,11, 20, (n+2) / n)==7/2

First fraction == 13 / 11

Note: The quadratic equation is:  5d^2 - 35d - 220 = 0, which gives d==11 and [d + 2] / d ==13 / 11

For this question we know the shortest altitude would come from the longest side which in this case can be either BC or AC because they both equal 20. Now we need to find the area of triangle ABC. For this question we should use a formula known as Heron's formula. Which basically says the area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) s = the semiperimeter or (a+b+c)/2 and a,b,c are the three sides of a triangle. Using heron's formula we get the semiperimeter = 55/2. \(\sqrt{55/2(55/2-20)(55/2-20)(55/2-15)}\) this simplifies to \(\sqrt{55/2(15/2)(15/2)(25/2)} = \sqrt{(15^2*55*25)/2^4}\) the denominator when square rooted equals 4. 15^2 goes out to 15 and 25 goes out to 5 leaving 55. Giving us the area of ABC as \(75\sqrt{55}/4\). Assigning the altitude as x. We get (x*20)/2 = \(75\sqrt{55}/4\). Which gets us our final answer of:

\(x = 75\sqrt{55}/40\)

Let the first fraction be ba​. We are told that a>b and a=b+2. Therefore, we know that a=b+2=a/b​→b^22−ab−2b+4=0→(b−2)(b−1)=0. Therefore, b=1 or b=2. We know that b must be an integer, so b=2. Therefore, the first fraction is 4/2.​

Could you give us the diagram?