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Hint:

use the Euclidean algorithm to find \(\gcd(4\cdot\frac{n(n+1)}{2}, n-2)\)

I do not think that there is anything wrong with Geno's answer.  Thanks Geno.

\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = 2^{-m} \int_0^{\frac{\pi}{2}} \cos^m x \, dx

\(\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx = 2^{-m} \int_0^{\frac{\pi}{2}} \cos^m x \, dx \)

\(LHS\\=\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx\\ =\int_0^{\frac{\pi}{2}} (\cos x \sin x)^m \, dx \\ =2^{-m}\int_0^{\frac{\pi}{2}} (2\cos x \sin x)^m \, dx \\ =2^{-m}\int_0^{\frac{\pi}{2}} (sin(2x))^m \, dx \\ \qquad let \;\;g=2x\\ \qquad when\;\;x=0,\;g=0\qquad when\;\;x=\frac{\pi}{2}\;\;g=\pi\\ =2^{-m}\int_0^{\pi } (sing)^m \, \frac{dx}{dg}\;dg \\ =2^{-m}\int_0^{\pi } (sing)^m \, \frac{1}{2}\;dg \\ \)

Below is perhaps less formal.

It can be deduced by knowing what the cos and sin graphs look like

and knowing they are the same except there is a phase shift of pi/2 radians

\(=2^{-m}\int_{\frac{\pi}{2}} ^{\pi } (sing)^m \;dg\\ =2^{-m}\int_{0} ^{\frac{\pi}{2}} (cosg)^m \;dg\\ =2^{-m}\int_{0} ^{\frac{\pi}{2}} (cosx)^m \;dx\\ =RHS\)

QED

LaTex:

LHS\\=\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x \, dx\\
=\int_0^{\frac{\pi}{2}} (\cos x \sin x)^m \, dx \\
=2^{-m}\int_0^{\frac{\pi}{2}} (2\cos x \sin x)^m \, dx \\
=2^{-m}\int_0^{\frac{\pi}{2}} (sin(2x))^m \, dx \\
\qquad let \;\;g=2x\\
\qquad when\;\;x=0,\;g=0\qquad when\;\;x=\frac{\pi}{2}\;\;g=\pi\\
=2^{-m}\int_0^{\pi } (sing)^m \, \frac{dx}{dg}\;dg \\
=2^{-m}\int_0^{\pi } (sing)^m \, \frac{1}{2}\;d

first, re-write the equation in LaTeX.

\({1\over2}y+{1\over4}=3+\boxed{\phantom{400}}y\)

Instead of a box, substitute it for x.

\({1\over2}y+{1\over4}=3+xy\)

To get rid of the fractions, multiply both sides by 4.

\(4*({1\over2}y+{1\over4})=(3+xy)*4\)

simplified...

\(2y+1=12+4xy\)

move all numbers to one side

\((2y+1)-(2y+1)=12+4xy-(2y+1)\)

simplified...

\(4xy-2y+11=0\)

the easiest way to get more than one solution is to make the equation a binomial.

to make \(4xy-2y+11=0\) a binomial, we can just say that x=y

simplified...

\(4y^2-2y+11=0\)

since x=y, y must be placed in the box.

The only thing I am worried about is that it might not want a variable for the box, but instead an actual number.

if $185=100% of the price, and a 5% markup is added, we add 5% to the total %.
105% is the same as 1.05x.

1.05*185=$194.25

The line y = 1 is a horizontal line that is one above the x-axis.

Since the x-coordinate determines how far left or right a point is, it stays the same at y=1.

To reflect a point is to mirror it accros a line. Since -5 is 6 away from 1, the reflected point must also be 6 away from 1. That point is 7.

Putting the coordinates together, we get the coordinate (6, 7), of course with the x-coordinate stated first.

We can start witht the number 500 and count all of the squares/cubes under that.

22² (484) is the closest we can get to 500 without going over it.
7³ (343) os the closest we can get to 500 without going over it.

22+7 is 29, which is how many we got rid of (almost, but not really)

Since 1 is both a square and a cube, we need to subtract 1 for the double count. 
Since 64 is both a square and a cube, we need to subtract another 1 for the double count.

So to compensate for each square and cube not included, we can add 27 (29-2) to 500
The 500th term would be 527, but 512 is 8^3, so we also have to compensate for that.

527+1=528

Look at this coincidence.  Your screen name tinfoilhat, or, if I may, "tin" as a nickname.  I answer questions on an eBay forum – https://community.ebay.com/t5/Ask-a-Mentor/bd-p/ask-a-mentor (btw my username on eBay is gosimus) – and the most amazing responder over there, username is tenbigdogs.  So, here it's "tin" and over there it's "ten" so is it really a coincidence... I think not. 

_.

Here is a clue:

pq-2p-2q-2=0

p(q-2)-2q-2=0

p(q-2)-2q+4 = 6

p(q-2)-2(q-2)=6

(p-2)(q-2)=6

Factors of 6: ±1,2,3,6

According to these factors, (p,q) = (1, -4), (3, 8), (4, 5), (5, 4), (8, 3)

5 favorable / 10 possible

1/2

Middle School method
Sean kept 1/5 of his stamps. So the rest is

5 - 5 * 1/5 = 4.

And the rest was given to Joe and Alex in

ratio of 3:5. So Joe got 4 * \( {3 \over 5 + 3}\) = 3/2; Alex

got 4 * \( {5 \over 3 + 5}\) = 5/2.

So the new ratio of Joe and Alex is (3 + 3/2):

(8 + 5/2) = 4 1/2 : 10 1/2.

Alex had 96 stamps more than Joe. So 96 / (10 1/2 - 4 1/2) = 96 / 6 = 16 stamps. (The "I" in ratio represent 16 stamps)

So Joe had 4 1/2 * 16 = 72 stamps.

(-2, 6) (-y, -12)

x, y,      x₂   y₂

L: y - y, = m (x - x,)

m = (y₂ - y₁)_/ x₂ - x₁ = (-12 - 6)/-4-(-2) = -18/-2 = 9

L: y - 6 = 9 (x - _-2)

L: y - 6 = 9x + 18

L: y = 9x + 24

x = 0

y = 9(0)+24

y = 24

The right angle sign denotes a 90˚ angle. This means that 6x+4 + 4x+6 = 90, so 10x+10 = 90 and x = ...

p and q are two straight lines. The vertically positioned angles they make are called vertical angles, and so are the horizontally opposite ones. What do you notice about these two vertical angles, and about the sum of 30 and x?

What do you notice about angle DCA and CAB? What are the sum of the angles in a triangle?

Have you heard of alternate interior/exterior angles?

Based on the information provided: P(x) = x^4 + ax^3 + bx^2 + cx + d

P(1) = 1 + a + b + c + d

After a bit of educated guessing, it seems this polynomial would be \(\left(x + \sqrt{1 + \sqrt6}\right) \left(x + \sqrt{1 - \sqrt6} \right) \left(x - \sqrt{1 + \sqrt{6}}\right) \left(x - \sqrt{1 - \sqrt6}\right)\)

Which simplifies to \(x^4-2x^2-5\)

I'm not entirely sure why this works, but I think it's because it can be turned into a quadratic by setting y=x^2 (and the square roots must cancel out). Anyone?

Aside from that, 1-2-5 = -6.

(a) Because collected $1680. 2/5 of the items sold were the bags

So the bags sold 1680 * 2/5 = $672

We can suppose the first day sold x bags 

So  x * 21 * 2/3 = 672 So x = 48

(b) We can suppose originally have y bags before selling 

So y * 1/3 = 48 So y = 144

So the items left 144 - 48 = 96

I recommend drawing a diagram.

I'm not completely confident about my answer, but here's what I got.

We know that DF+FC+CE+EB = 124/2 = 62, and that 20(DF + FC) = 7(CE + EB), which are just 2 different ways to calculate the area of the parallelogram.

We need to calculate either 20(DF + FC) or 7(CE + EB). 

From our first equation, we get DF + FC = 62 - (CE + EB), so 20(62-(CE+EB)) = 7(CE+EB). Solving, CE+EB = 1240/27 and so the area is 7 * 1240/27 = 8680/27.

See https://web2.0calc.com/questions/help-geometry_74

Blue circle radius = 0.154700538

Red circle radius = 0.06278172

That's all that I have so far. 

It seems that the same question has been asked on brainly.com (https://brainly.com/question/12971062), and the solution matched your answer of $3.49. 

Expand everything in base 10 form. 1000A + 100B + 10C + D + 100B + 10C + D = 2012.

1000A + 200B + 20C + 2D = 2012

Divide by 2.

500A + 100B + 10C + D = 1006.

D has to be 6.

500A + 100B + 10C = 1000

Divide by 10.

50A + 10B + C = 100

C has to be 0.

50A + 10B = 100

5A + B = 10

(A,B) = (1,5) or (2,0)

Solutions: ABCD = 1506 or 2006.

We cannot have a solution without a 0, so there is no solution. 

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In case you mean A*B*C*D + B*C*D = 2012:

B*C*D(A+1) = 2012

One of these must include 503 (prime), which is not a digit. So this is not it either.

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I found another question that has a realistic solution at https://brainly.com/question/17335668, is this perhaps what you meant?

Without loss of generality, say Benny has $120.

He just spent 24 dollars on 4 pineapples and 8 mangoes.

Since a pineapple is 4 times a mango, Benny spent 1 dollar per mango and 4 dollars per pineapple. 

He now has 96 dollars, and he spends $48 on pineapples, accounting for 12 pineapples. 

He has bought 12+4 = 16 pineapples.

" such that AK in altitude to ABC."

I'm assuming this means that AK is an altitude to BC. I suggest drawing a diagram.

If CK = 4 and AC = 9, then by the Pythagorean Theorem, AK^2 = 9^2-4^2 = 65, so AK = √65. Since BK is √5, AB^2 = 65 + 5 = 70, so AB = √70.