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They are equal. If you divide the diagram into equilateral triangles, there are 12 of each color.

hi and welcome to the site! :)
i have an account on aops, and I'm pretty sure there's a lot of ppl here who have an account there too since 90% of all questions on this forum are aops questions lol

please fix your code! :)

hi guest!

so if 1 tower is 18m and another is 30m it would form a trapezoid, something like this:

*this is not drawn to scale*

as you can see, it forms a right triangle, with legs of 16 and 30-18=12!

so, the missing length is \(16^2+12^2=y^2\)

\(400=y^2\)

\(\boxed{y=20}\)

i hope this helped you! :)

hey guest!
 

so we can express x as 2y. so the first equation becomes 2y+1=y-8.

2y+1=y-8

2y+9=y

y+9=0

y=-9

I'm sorry but how would I get the simplified version into the slope-intercept form?

Every factorial from 6! and up all the way to 100!

Count how many 1*1, 2*2, 3*3... n*n squares there are. Hope that helps. 

Thanks a lot! 

So the answer would be \(9889\) 

PROBABILITY has 11 letters.

The most straightforward thought would be the factorial of 11. But there are repeating letters, so we would have double-counted if we say the number of ways is 11!.

In fact, we need to divide by the factorial of the number of each type of repeating letters.

The number of ways would be \(\dfrac{11!}{\color{green}2!\color{blue}2!\color{black}} = 9979200\)

The green one is for repeating 'B's, and the blue one is for repeating 'I's.

We can parameterize the coordinates of P.

\(P = (t, 5t + 3)\) for \(t\in\mathbb R\)

We use the formula to calculate the midpoint of PQ.

\(M = \left(\dfrac{t + 3}{2}, \dfrac{5t + 3 - 2}{2}\right) = \left(\dfrac{t + 3}{2}, \dfrac{5t +1}{2}\right)\)

Then, we let M_x be the x-coordinate of M and M_y be the y-coordinate of M. 

We then find the relationship between M_x and M_y.

Consider 10M_x.

\(10M_x = 5(t + 3) = 5t + 15 = (5t + 1) + 14 = 2M_y + 14\)

\(10M_x - 2M_y - 14 = 0\\ 5M_x - M_y - 7 = 0\)

Therefore the point M must lie on the line \(5x - y - 7 = 0\)

Have you heard of the "inverse Pythagorean theorem"?

Let a and b be the legs of a right triangle, and h be the altitude to the hypotenuse.

The inverse Pythagorean theorem states that \(h^{-2} = a^{-2} + b^{-2}\).

We can use this equation to solve this problem.

Let h be the altitude to the hypotenuse.

\(\dfrac1{h^2} = \dfrac1{8^2} + \dfrac1{15^2}\\ h^2 = \dfrac{120^2}{8^2 + 15^2} = \left(\dfrac{120}{17}\right)^2\\ h = \dfrac{120}{17}\)

Assuming \(x_y\) means "x in base y".

The equation given implies 

\(7\Diamond + 4 = 8\Diamond + 1\)

The above result is obtained by changing both sides to base 10.

Solving, we have \(\Diamond = 3\).

Let \(H\in AB\) such that \(PH \perp AB\) and let \(T\in CD\) such that \(PT \perp CD\)

\(\dfrac{AB}2 = \dfrac{AB\cdot PH}{2}\\ PH = 1\)

Let x be the side length of the square and considering similar triangles,

\(x = \dfrac1{x - 1}\)

I will leave the solving process for you as an exercise.

At the end, we get a beautiful result, which is \(x = \varphi\), where \(\varphi\) is the golden ratio.

Let BC = a, AC = b.

Now, \(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)

and \(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)

According to the question, AM = 19 and BN = 13.

\(4b^2 + a^2 = 38^2 \\4a^2 + b^2 = 26^2\)

Adding gives

\(a^2 + b^2 = \dfrac{38^2 + 26^2}{5}\)

The length of the hypotenuse is \(\sqrt{a^2+ b^2} = 2\sqrt{106}\)

Let AZ = a, ZW = b.

We know that \(b^2 - a^2 = 273\)

The yellow area, using the formula for area of trapezoid, is \(\dfrac{(a + b)(b - a)}2\).

You can continue from here. This is just one step away from the answer.

‖a⃗‖ = \(\sqrt{(4 - (-2))^2 + (-2 - (-5))^2} = \sqrt{6^2 + 3^2} = 3\sqrt 5\)

Let h(x) be the function you need.

From the graph, you see h(4) = f(1). Knowing that f is translated horizontally to h, h(x) = f(x - 3).

\(\mathcal P(\text{same color}) = \dfrac29\cdot\dfrac18 + \dfrac39\cdot \dfrac28 + \dfrac49\cdot\dfrac38 = \dfrac5{18}\)

We can actually solve the equation to see how many real solutions there are.

\(\begin{cases}y = x^2 - 5\\x^2 + y^2 = 25\end{cases}\\ x^2 + (x^2 - 5)^2 = 25\\ x^4 - 9x^2 = 0\\ x^2 = 0 \text{ or } x^2 = 9\\ x = -3, 0, 3\)

Plugging in these values into the equation gives

\(y = 4, -5, 4\)

in this order.

Therefore there are 3 pairs of real solution, namely \((-3, 4), (0, -5), \text{ and }(3, 4)\)

Convert from molecules into milliliters:
3.27*10^25 molecules of[H, 2]O (water)

Divide the number of moles by AN to convert to moles:
(3.27*10^25 molecules)/1 * (1 mol)/(6.022*10^23 molecules) == 54.3 mol

The amount-of-substance, n, is:
n == 54.3 mol

m == n M == (54.3 mol)/1 * (18.01528 g)/(1 mol) == 978.21 g

Convert from grams to cubic centimeters using the relation 1 (cm)^3  (cubic centimeter) == 1 mL  (milliliter)[[2]] and the density 1 g/(cm)^3  (gram per cubic centimeter) at STP of [H, 2]O:
(978.21 g)/1 * (1 (cm)^3)/(1 g) == 978.21 (cm)^3

Convert to milliliters using the relation 1 (cm)^3  (cubic centimeter) == 1 mL  (milliliter):
(978.21 (cm)^3)/1 * (1 mL)/(1 (cm)^3) == 978.21 mL

You can use the definition of derivative on \(f'(x)\) to find \(f''(x)\) because \(f''(x) = \displaystyle\lim_{h\to0}\dfrac{f'(x + h) - f'(x)}{h}\).

If f is increasing on an interval \(\mathcal I\), then for all real numbers \(x\in \mathcal I\), \(f'(x) \geqslant 0\).

If f is decreasing on an interval \(\mathcal I\), then for all real numbers \(x\in \mathcal I\), \(f'(x) \leqslant 0\).

If f is concave downward on an interval \(\mathcal I\), then for all real numbers \(x\in \mathcal I\), \(f''(x) \leqslant 0\).

If f is concave upward on an interval \(\mathcal I\), then for all real numbers \(x\in \mathcal I\), \(f''(x) \geqslant 0\).

These are the definitions. I believe you can finish the rest with these.

The area is equal to \(\left|\mathbf{u} \times \mathbf{v}\right|\).

Notice that \(\mathbf{u}\times \mathbf{v} = \det\begin{bmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&-2&0\\6&-2&0\end{bmatrix} = \det\begin{bmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&-2&0\\1&0&0\end{bmatrix} = 2\mathbf{k}\)

Therefore \(\left|\mathbf{u} \times \mathbf{v}\right| = |2\mathbf{k}| = 2\)

Therefore the area of the required parallelogram is 2 sq. units.

Yes.