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θ=5π/6

Welp looks like I got it right. I guess I just misread the question.

Thank you so much!

Hello Guest!

When the particle visits the point \((\frac{3}{2},\frac{3}{2})\) we get the following equations:

\(\dfrac{3}{2}=1+sin(\pi t) \iff sin(\pi t)=\dfrac{1}{2} \space \space \space \space \space \space[Eq.\space 1] \\ \space \\ \dfrac{3}{2}=3sin(\pi t) \iff sin(\pi t) = \frac{1}{2} \space \space \space \space \space \space [Eq. \space 2]\) 

They are the same equation, so we need to solve:

\(sin(\pi t) =\dfrac{1}{2}\), and we want the first four positive solutions of t.

Lets solve for t:

\(t=\dfrac{1}{\pi}sin^{-1}(\dfrac{1}{2})\)

Now, \(sin^{-1}(\dfrac{1}{2})=30,150,390,510,750,870,...\) (degrees) and so on. (30 and 150 in first and second quadrant in the unit circle and then we keep adding 360 on each of them to get infinite solutions). However, we only need the first four, and since the question is in radians so we have to use radians.

Therefore,

 \(t_1=\dfrac{1}{\pi}(\dfrac{\pi}{6})=\dfrac{1}{6}\)

\(t_2=\dfrac{1}{\pi}(\dfrac{5\pi}{6})=\dfrac{5}{6}\)

\(t_3=\dfrac{1}{\pi}(\dfrac{13\pi}{6})=\dfrac{13}{6}\)

\(t_4=\dfrac{1}{\pi}(\dfrac{17\pi}{6})=\dfrac{17}{6}\)

I hope this helps!

So I was looking through it, and I found that CPhill already answered this question, but it is wrong. Can someone please help

Thank you in advance. 

-2 and 4

4/5 * 1/7 = 4/35 of a gallon

Oh! I am sorry, I distributed the bracket wrong in the first answer (at -(3sqrt(z) -1) I make this: -3sqrt(z)+3 while it should have been -3sqrt(z)+1.

Ok here is the correct way: 

\(\dfrac{2\sqrt{z}-3-\sqrt{z}+1}{\sqrt{z}-1}=-\dfrac{3\sqrt{z}-1}{\sqrt{z}-1}\)

So:

\(\frac{\sqrt{z}-2}{1}=-3\sqrt{z}+1 \implies \sqrt{z}=-3\sqrt{z}+3 \implies 4\sqrt{z}=3 \\ \sqrt{z}=\frac{3}{4} \\ z=\frac{9}{16}\)

I'm sorry but it says that's wrong. Still, thanks for trying!

Hi Fytonic!

Let's simplify the left-hand side first:
\(\frac{2\sqrt{z}-3-\sqrt{z}+1}{\sqrt{z}-1}=\frac{\sqrt{z}-2}{\sqrt{z}-1}\)

Now, let's equate this to the right-hand side as given:

\(\frac{\sqrt{z}-2}{\sqrt{z}-1}=\frac{3\sqrt{z}-1}{1-\sqrt{z}} \)

Now, notice: \(\frac{\sqrt{z}-2}{\sqrt{z}-1}=\frac{3\sqrt{z}-1}{-(-1+\sqrt{z})} =-\frac{3\sqrt{z}-1}{\sqrt{z}-1}\)   (we factored a negative from the denominator). So when we multiply both sides by: \(\sqrt{z}-1\), there is no longer a denominator:

\(\implies \sqrt{z}-2=-(3\sqrt{z}-1)\)

\(\implies \sqrt{z}-2=-3\sqrt{z}+3 \\ \implies 4\sqrt{z}=5 \\ \implies \sqrt{z}=\frac{5}{4} \\ \implies z=(\frac{5}{4})^2=\frac{25}{16}\)

I hope this helps!

https://artofproblemsolving.com/texer/xjsiitim

Question properly displayed here. Click "With bbcode"

https://math.stackexchange.com/questions/3825841/sherlock-holmes-base-number-problem

If we plot the given info in a Venn diagram, it would look like this: 

We will first solve for the number of people who only play cricket. 

Since we know the total is 55, we can write an equation: \(c + 13 + 3 + 25 = 55\), meaning \(c =14\)

Next, we know that \(b + x + 16 = 35\) and \(f + x + 28 = 53\). 

But, knowing that there are 90 students in total, we can write the equation \(b + x + f = 90\). 

Now, just solve for f. Hint: Simplify the equations and substitute 1 into another

maybe format your question properly?

If a quadratic has a repeated root, the discriminant (\(b^2 - 4ac\)) is equal to 0. 

Can you take it from here? Feel free to ask if you need more help

\(5^n + 8^n + 13^n + 17^n \mod6\\ \equiv (-1)^n + 2^n + 1^n + (-1)^n \mod6\\ \equiv 2*(-1)^n + 2^n + 1 \mod6\\\)

\(\qquad 2^1 \mod6 \equiv 2 \mod 6\\ \qquad 2^2 \mod6 \equiv 4 \mod 6\\ \qquad 2^3 \mod6 \equiv 2*2 \mod 6\equiv 4 \mod6\\ \qquad 2^4 \mod6 \equiv 4*2 \mod 6\equiv 2 \mod6\\ \qquad etc\qquad so\\ \qquad 2^n \equiv 2\mod 6\qquad \text{If n is odd}\\ \qquad 2^n \equiv 4\mod 6\qquad \text{If n is even}\\ \)

\(5^n + 8^n + 13^n + 17^n \mod6\\ \equiv (-1)^n + 2^n + 1^n + (-1)^n \mod6\\ \equiv 2*(-1)^n + 2^n + 1  \mod6\\ \text{If n is odd}\\ \qquad \equiv 2*(-1) + 2 + 1  \mod6\\ \qquad \equiv -2 + 2 + 1  \mod6\\ \qquad \equiv 1  \mod6\\ \text{If n is even}\\ \qquad \equiv 2*(1) + 4 + 1  \mod6\\ \qquad \equiv 2 + 4 + 1  \mod6\\ \qquad \equiv 1  \mod6\\ so\\ \text{For all positive integer values of n}\\ 5^n + 8^n + 13^n + 17^n \equiv 1 \mod6\\ \text{It will never have a zero remainder when divided by 6}\)

LaTex:

5^n + 8^n + 13^n + 17^n \mod6\\
\equiv (-1)^n + 2^n + 1^n + (-1)^n \mod6\\
\equiv 2*(-1)^n + 2^n + 1  \mod6\\
\text{If n is odd}\\
\qquad \equiv 2*(-1) + 2 + 1  \mod6\\
\qquad \equiv -2 + 2 + 1  \mod6\\
\qquad \equiv 1  \mod6\\
\text{If n is even}\\
\qquad \equiv 2*(1) + 4 + 1  \mod6\\
\qquad \equiv 2 + 4 + 1  \mod6\\
\qquad \equiv 1  \mod6\\
so\\
\text{For all positive integer values of n}\\
5^n + 8^n + 13^n + 17^n \equiv 1 \mod6\\
\text{It will never have a zero remainder when divided by 6}

Hi Guest!
First, let's draw the triangle and all of the given:

Now, to find the area of triangle AEC, we already have the base (AC=30), we need the height, that is, the distance from E to AC. 

Notice: AM=22 (44/2, as M is the midpoint of AB). and ME is given to be 10.

So with trigonometry: \(tan(\theta)=\frac{10}{22}=\frac{5}{11}\)

And, by pythagoras: \(AE=\sqrt{22^2+10^2}=\sqrt{584}\)

If \(tan(\theta)=\frac{5}{11} \implies sin(\theta)=\frac{5}{\sqrt{146}}\) (Or simply using the triangle above, sin(theta) = opposite/hypotenuse which is ME/AE).

Next, in the triangle with the "Other theta" That is, the triangle having "h" 

\(sin(\theta)=\frac{h}{\sqrt{584}} \iff \frac{5}{\sqrt{146}}=\frac{h}{\sqrt{584}}\)

Therefore, \(h=10\)

So, the area of triangle ACE = \(\frac{30*10}{2}=150\)

:D        thankss

\(\frac{1}{5^1} + \frac{1}{5^2} + \frac{1}{5^3} + ...\)

The expression above can be determined as the sum of an infinite geometric series

Each term can be written as \(1*(\frac{1}{5})^n\)

Using this information, we can plug the values into the formula for the sum of an infinite geometric series

\(\frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{1}{4}\)

Finally we get an answer of 1/4

:D

yes please

To figure out whether a fraction is a terminating decimal, the denominator of the fraction must be a power of 2 and/or 5 ONLY. We know 1+ 2x won't be a multiple of 2 as x is an integer. We can however express the denominators as powers of 5.

So all we have to find is:

\(\frac{1}{5^1} + \frac{1}{5^2} + \frac{1}{5^3} + ...\)

If you need more help, respond to this answer I guess.

For those confused, the addition problem was SEAS + EBB + SEA = BASS