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By sum-to-product formula:

\(\sin \dfrac x2 + \sin \dfrac x3 = 2 \sin \dfrac{5x}{12} \cos\dfrac{x}{12}\)

The function \(g(x) = \sin \dfrac{5x}{12}\) has period \(\dfrac{24\pi}5\), which means \(g\left(\dfrac{24n\pi}5 + x\right) = g(x), n\in \mathbb Z\)

The function \(h(x) = \cos \dfrac{x}{12} \) has period 24 pi, which means \(h(24n\pi + x) = h(x), n\in \mathbb Z\)

Therefore \(f(24\pi + x) = 2g(24\pi + x) h(24\pi + x) = 2g\left(\dfrac{24\pi}5 \cdot 5 + x\right)h(24\pi \cdot 1 + x) = 2g(x) h(x) = f(x)\)

Therefore the period is \(24\pi\).

No smaller T satisfies \(f(T + x) = f(x)\)

I tried answering the question before I went here and put in 3 and it was wrong.

expected value is a predicted value of a variable, calculated as the sum of all possible values each multiplied by the probability of its occurrence.

\((a + b)^2 = (a - b)^2\\ (a + b)^2 - (a - b)^2 = 0\\ \text{Using }a^2 - b^2 = (a - b)(a + b),\\ (2a)(2b) = 0\\ ab = 0\\ a = 0\text{ or } b = 0\)

AA9_{12} is right.

The WolframAlpha computation: https://www.wolframalpha.com/input/?i=14A6+base+12+-+5B9+base+12

Assuming you meant \(y = \dfrac25 x\),

Substituting (p, q) into the equation, \(q = \dfrac{2p}5\),

Because pq = 90, p(2p/5) = 90, which means p² = 225

Taking square roots on both sides, p = 15.

There can only be one even digit...so the other three are odd. Let O represent an odd digit and E an even digit.

The numbers can be like this:

EOOO

OEOO

OOEO

OOOE

In each number, there are 5 possible odd digits(1, 3, 5, 7, 9) and 5 possible even digits(0, 2, 4, 6, 8). In the first case, EOOO, the first digit can't be 0 so there are 4*5*5*5=500 ways. 

The other three cases are the same: each one has 5^4=625 ways.

So the total number of integers with only one even digit is 624+500=1125.

Let me know if this is correct.

This has been posted few days ago.

https://web2.0calc.com/questions/pls-help-due-today_1

We can rewrite this as 

\(\begin{cases}\log_w m = \dfrac1{12}\\\log_w n = \dfrac1{16}\\\log_w p = \dfrac1{36}\\\log_{w} mnpq = \dfrac1{72}\end{cases}\)

By log properties, 

\(\log_w m + \log_w n + \log_w p + \log_w q = \dfrac1{72}\\ \dfrac1{12} + \dfrac1{16} + \dfrac1{36} + \dfrac1{\log_q w} = \dfrac1{72}\)

The rest is trivial.

Ans: \(\log_q w = -\dfrac{144}{23}\)

Hint: Consider the ratio of side lengths of each triangle.

Oh yes, I forgot that.

Hello! I am new to this website and I am not the best at math, but I like helping people so I will try to help.  Please tell me if I get it wrong, and don't be mad. Thanks!

My thinking is that there are exactly five possible even digits, and we can first take a look at the possible 4-digit numbers where the first digit is even.  There are 4 choices for the first one because it can't be 0, and 10 for each one following.  This gives 4*10*10*10=4000.  We can do the same for when the even digit is in the hundereds, tens, and ones place, then add them all up: 4000+5000+5000+5000=19000.  So the answer is 19000.  Again, I might not be correct. :)

\(f(x) = (x^2 + 6x + 9)^{50} - 4x + 3 = (x + 3)^{100} - 4x + 3\)

Substituting \(r_1, r_2,\cdots,r_{100}\) into f(x) = 0, we have

\(\begin{cases}\left(r_1 + 3\right)^{100} - 4r_1 + 3 = 0\\\left(r_2 + 3\right)^{100} - 4r_2 + 3 = 0\\\cdots\\\left(r_{100} + 3\right)^{100} - 4r_{100} + 3 = 0\end{cases}\)

Moving all the terms to the right hand side and adding them up,

\((r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100} = 4(r_1 + r_2 + \cdots + r_{100}) - 300\)

That means the required answer is 4(sum of roots) - 300.

By Vieta's formula, \(\text{sum of roots} = -\dfrac{\text{coefficient of }x^{99}}{\text{coefficient of }x^{100}}\).

Using binomial theorem, the coefficient of x¹⁰⁰ is 1 and the coefficient of x⁹⁹ is 300.

Therefore \(r_1 + r_2 + \cdots + r_{100} = -300\).

Therefore \((r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100} = 4(-300) - 300 = \boxed{-1500}\)

Let t = 1/x.

\(25^{-2} = \dfrac{5^{48t}}{5^{26t} \cdot 25^{17t}}\)

Rewrite 25^n into 5^(2n),

\(5^{-4} = \dfrac{5^{48t}}{5^{26t}\cdot 5^{34t}}\)

By law of indices,

\(5^{-4}= 5^{-12t}\)

Equating the powers would give t = 1/3, which means x = 3.

The bag of loonies are worth $400 in total ⇔ There are 400 loonies in the bag.

But one loonie = 4 dimes in terms of mass, so if the bag of dimes has the same mass as a bag of loonies, then there should be (4)(400) dimes, which is 1600 dimes.

The bag of dimes is worth $1600 (0.10) = $160.

Notice that \(n^5 + n^4 + 1 = (n^2 + n + 1)(n^3 - n + 1)\).

That means, for all \(n^2 + n + 1 \neq 1\) and \(n^3 - n + 1 \neq 1\), \(n^5 + n^4 + 1\) is composite.

This means when \(n^5 + n^4 + 1\) is prime, \(n^2 + n = 0\) or \(n^3 = n\)

Solving, we get n = -1 (rejected) or n = 0 (rejected) or n = 1.

When n = 1, \(n^5 + n^4 + 1 = 3\), which is a prime.

The only possibility is n = 1.

\(y' = \dfrac{y(t + 1) + (t + 1)^2}{t^2}\\ \dfrac{dy}{dt} = y\left(\dfrac1t + \dfrac1{t^2}\right) + \left(1 + \dfrac1t\right)^2\\ \dfrac{dy}{dt} - \left(\dfrac1t + \dfrac1{t^2}\right)y = \left(1 + \dfrac1t\right)^2\\ \)

We then find the integrating factor:

\(\mathcal{I}(t) = e^{-\int \left(\frac1t + \frac1{t^2}\right)\,dt} = e^{-\ln t + \frac1t} = \dfrac1t e^{1/t}\)

This means 

\((y\mathcal I)' = \left(1 + \dfrac1t\right)^2 \mathcal I\)

\(\dfrac{y}t e^{1/t} = \displaystyle \int \dfrac1t \left(1 + \dfrac1t\right)^2 e^{1/t}\,dt\)

You will need some special functions to evaluate this integral, so good luck. Hope this helps.

(√2 * x)² + x² = 20²

3x² = 400

x² = 400 / 3

x = 11.547   

No, (5, 5) is also possible.

NVM it's 13/2

∠TAC = ∠TAM

cos (∠TAC)  = ( 11.4 / 2 ) / 12

∠TAC = 61.6º  

This answer would give you 5/5 marks.  

I'm pretty sure that's correct but I think one of your steps are wrong...

TOTAL SURFACE AREA OF THE CYLINDER(TSA)=462CM²
CURVED SURFACE=1/3 OF TSA
=1/3 X 462
=154CM²
 AREA OF EACH CIRCLE=(462-154)/2
=154CM²
PI*R²=154
R²=49
R=7
2PI*R*H=154
H=154*7/22*7
H=3.5
VOLUME=PI*R²*H=22/7*49*35/10
VOLUME=539CM³

After solving the problem, I'm not really sure how this hint ties into the problem. 

Are you saying there is 1 option for a, 2 options for b, 3 options for c, and 6 options for b? Or did you factorize 36? 

Just curious to see if your solution was maybe faster than casework.