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The octahedron currently has 6 vertices. Each removal of a square-based pyramid removes one vertex and adds 4 more; this occurs 6 times. Hence we end up with 4 times 6= 24 vertices.

f(x) is the function, so here, all you need to do is put in x=7. 

(5*7 +1)/(7-1)

=36/6

=6

f(7)=6

Hope this was helpful.

At first, they had the same amount. I will call this x.

Marcus now has 3/4 *x, and Kevin has 2/3 *x. 

Marcus has 15 more than kevin, so I can list this as 3/4 *x=2/3 *x +15

1/12 *x=15

x=15*12

x=180.

Marcus gave away 1/4, so I multiply 180 and 1/4, which gets me 45. 

Marcus gave away 45 Stamps.

That's the WRONG answer. It's easy to get wrong answers but hard to get correct answers

Let me express this in algebra form. 

Lets put Quail as Q, and Chicken as C.

3C=2Q

3C-210/2Q = 1/3

Lets bring 2Q into 3C.

2Q-210/2Q=1/3

We can use cross multiplication.

6Q-630=2Q

4Q=630

Q=157.5

2Q=315

There are 315 Quail eggs.

But, we want the total number of eggs at the end, so we must find the amount of chicken eggs too.

3C=2Q

3C=315

We want 3C-210.

315-210=105

So the total number of eggs is 105+315=420 eggs at the end.

That's what I was thinking too lol but I thought it was a bit hard to type it out. Pretty complex.

The maximum value is n.

a_{12} = 10*sqrt(3)/3.

Radius = 4√13

By Cauchy-Scwharz, the minimum value is 1/(n + 1).

By AM-GM, the maximum value is 3/4.

In the class library, 3/5 of the books are fiction books. 2/3 of the remaining books are non-fiction books. The rest are magazines. There are 10 fewer magazines than non-fiction books in the class library. How many fiction books are there altogether?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

x    => all books and magazines

3x/5   => fiction books

2/3(x - 3x/5)    => non fiction books

2/3(x - 3x/5) - 10    => magazines

I'm sure you can find your answer. 

Let me express this using algebra. Vernon is V, Walsh is W, and the number of staps sold per person is X.

V= 2/5W which is 5/2V= W

V-X=57

W-X=57*3

Let me bring V= 2/5W into W-X=57*3 first.

5/2V-X=57*3

We can subtract V-X=57 from 5/2V-X=57*3

3/2V=57*3 -57

3/2V=114

V=76

Okay, almost done!

Let's bring V=76 into 5/2V= W

So W= 190 and V=76

Any questions? I hope this helped.

Let's write it out using algebra. Iskander is I, Anthony is A, and Sean is S.

I+A+S=555

A= I+13 

S=(A+I)2

Let's solve for I first.

Lets bring A= I+13 into S=(A+I)2

S=((I+13)+I)2

S=(2l+13)2

S=4I+26

Lets bring S=4I+26 and A= I+13 into I+A+S=555

I+I+13+4l+26=555

6I=516

I=86

But, we want A. So we bring I=86 into A= I+13.

A=86+13

A=99

Hope this helps.

942! = 2^935 * 3^467 * 5^233 * 7^155.......etc.

80 ==2^4  x  5.  Since the exponent of 5 ==233  and 80==2^4  x  5, it therefore follows that the largest possible value of n ==233. So that:

942! mod 80^233 ==0 

Using desmos, the circle you said will go through the point (6, 0). The shortest distance will be perpendicular to the line you said, and passes through the point (6, 0). Let's put the equation 3x + 4y = 25 into y = mx + b form. ==> 4y + 3x = 25 ==> 4y = -3x + 25 ==> y = (-3/4)x + 25/4. A line perpendicular it would have slope 4/3. Using point slope form, the equation perpendicular to 3x + 4y = 25 would be y - 0 = 4/3(x - 6) ===> y = (4/3)x - 8.  The point the equations y = (-3/4)x + 25/4 & y = (4/3)x - 8 will be (171/25, 28/25). The distance from (6, 0) to (171/25, 28/25) is 1.4. 

(ik this solution probably doesnt make sense)

When the denominator is 0, the fraction is undefined. the denominator in this problem is y^2 -5y +4 +y^2 -6y.

We want this to be 0, so I add that at the end.

y^2 -5y +4 +y^2 -6y=0

We can simplify this.

2y^2 -11y +4=0

Bring that into the quadratic equation, and then you should have the answer.

Hope this helps.

The general equation for any circle uses this:

(x-h)^2 + (y-k)^2 = r^2 with (h, k) being the center and r being the radius.

We already know what the center is, so now all we need to do is find the radius. we can easily find the radius with the pythagorean theorem.

We can imagine a right triangle with the radius from (2, 2) to (4, 5) as the hypotenuse.

Let's look for the legs first. 4-2=2. The second leg is 5-2=3.

Pythagorean theorem:

2^2 + 3^2 = square root of 13

r^2= 13

We can bring r^2 and (h,k) into the general equation.

The equation you are looking for is (x-2)^2 + (y-2)^2 = 13

I hope this helps.

First of all, this problem asks for integers, which means that you want the sum of integers from -12 to 13, inclusive. we can ignore the zero, because there is no value of that. Numbers -12 to 12 cancel out with their opposite number. this leaves one number, 13. The sum of the integers between -12.1 and 13.3 is just 13.

Suppose that f(x) and g(x) are functions on R such that the range of f is [-5,5], and the range of g is [-2,2]. The range of f(x) times g(x) is [a,b]. What is the largest possible value of b?

Hello Guest!

\(-5\leq R_f\leq 5\\ -2\leq R_g\leq 2\\ \)

\(h(x)=f(x)\cdot g(x)\)

The smallest value that h (x) can, but does not have to have, is \(-2\cdot 5=\color{blue}-10.\)

The largest value that h (x) can, but does not have to have, is \(2\cdot 5=\color{blue}10.\)

You need the text of f(x) and g(x) because the maximum or minimum of the two functions can each be on a different x value.

  !

I got 28%

But I have not double-checked my answer.

I think guest just made a careless mistake that is commonly made Ginger.

I do not think this error was intentional,.

Note (check mark notwithstanding) that the guest’s answer is absolute bullshit.

This was posted by the forum’s Wrong Answer Troll (WAT), who occasionally posts work product to give the appearance of legitimacy to his wrong answers.   

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The equation $\dfrac{a+b}{a-3b}=5$ is a line with a slope of $m= \dfrac{b}{a} \rightarrow \dfrac{1}{4}$

The inverse of this slope is the ratio of $\dfrac{a}{b} \rightarrow \dfrac{4}{1}$

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I’ve followed WAT for almost two years. A few weeks ago I zeroed in on a very likely suspect from my short list. (I can’t prove it, though). He’s a long-time member who periodically posts reasonably competent solutions to moderately advanced questions. There are two related sources for his primary motivation in creating these wrong answer posts. I’m not going to elaborate beyond adding that this derivation is based on the parameters of the question posts he trolls.

This forum, like the world at large, is full of fascinating people.  Most human-equations are probably not fully solvable, but close approximations are available statistically, with reasonable confidence intervals. 

One open question is how he will react to this post, assuming he sees it. There is always a reaction; it’s not always obvious. 

GA

In the equation 1/j + 1/k = 2/3, both j and k are positive integers. What is the sum of all possible values for k?

Hello Guest!

1/j + 1/k = 2/3

\( \frac{1}{k}=\frac{2}{3}-\frac{1}{j}\\ \frac{1}{k}=\frac{2j-3}{3j}\\ \color{blue}k=\frac{3j}{2j-3}\)

\(j\in\{\)  0,  1,  2,  3 \(\}\)

\(k\in\{\)  0, -2,  4,  2 \(\}\)

The sum of all possible values for k is 4.

  !