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Hello x^2 it's good to see you back after you were kinda inactive for a bit :)

An arithmetic sequence is a sequence of numbers such that the difference of consecutive terms is constant. One can find the nth term of an arithmetic sequence with \(a_n=a_1+d(n-1)\).

\(a_n\) represents the value of the nth term of the sequence.

\(a_1\) represents the value of the first term of the sequence.

\(d\) is the common difference, or the difference of consecutive terms

\(n\) represents the desired term, 10 in this case.

The common difference is 7, and the first term is 8, and we are looking for the 10th term of this sequence.

\(a_{10}=8+7(10-1)\\ a_{10}=8+70-7\\ a_{10}=71\)

This is a duplicate question, so I will link you to the URL where CPhill answered this question:

You have asked a lot about this topic lately, so I encourage you to view the other answers we all have provided as examples for your sketch for this one. I have done two already. I will provide the hyperlink to both below.

You have asked a lot about this topic lately, so I encourage you to view the other answers we all have provided as examples for your sketch for this one. I have done two already. I will provide the hyperlink to both below.

Here is a sketch:

\(\frac{4\pi}{3}\) lies between \(\pi\) and \(\frac{3\pi}{2}\), so the terminal side lands in the third quadrant.

This is a duplicate post, so the hyperlink below leads you to the URL where I answered this already.

What is the value of \(\sin\left(\frac{5\pi}{2}\right)\)?

Well, first, let's determine the quadrant or quadrantal where the terminal side lies. Rotating counterclockwise by \(\frac{5\pi}{2}\) radians exceeds one revolution, and the terminal side will land on the first quadrantal:

\(\sin(\frac{5\pi}{2})=\sin(\frac{\pi}{2})=1\)

Oops! Thank you, you are correct. I've kept a 5! in the denominator, after I'd cancelled it with part of the 10! in the numerator.  I'll change it.

Above is a diagram of the situation. We can use the formula \(L = r*\theta_{rad}\) where \(L\) is the arclength subtended by the central angle and \(r\) is the radius of the circle and \(\theta_{rad}\) is the measure of the central angle in radians.

You should be able to finish it from here.

Alan: Your above calculation is slightly out! : (10 nCr 2)*(8 nCr 3) =2520

As follows:

\(\frac{10!}{2!(10-2)!}\times \frac{8!}{3!(8-3)!}= \frac{10!}{2!}\times \frac{1}{3!5!}=\frac{10*9*8*7*6}{2*6}=2520\)

Edited to correct silly error!

If the 4014th term of a geometric sequence of non-negative numbers is 135,

and the 14th term is 375,

what is the 2014th term?

\(\begin{array}{|rcll|} \hline i < j < k \\ a_i &=& ar^{i-1} \\ a_j &=& ar^{j-1} \\ a_k &=& ar^{k-1} \\ \hline \dfrac{a_k}{a_i} &=& r^{(k-1)-(i-1)} \\ \mathbf{\dfrac{a_k}{a_i}} &=& \mathbf{r^{k-i}} \\\\ \dfrac{a_k}{a_j} &=& r^{(k-1)-(j-1)} \\ \mathbf{\dfrac{a_k}{a_j}} &=& \mathbf{r^{k-j}} \\ \hline r= \left(\dfrac{a_k}{a_i}\right)^{\frac{1}{k-i}} &=& \left(\dfrac{a_k}{a_j}\right)^{\frac{1}{k-j}} \\\\ a_k^{\frac{1}{k-i}} a_j^{\frac{1}{k-j}} &=& a_k^{\frac{1}{k-j}} a_i^{\frac{1}{k-i}} \\\\ a_j^{\frac{1}{k-j}} &=& a_k^{\frac{1}{k-j}-\frac{1}{k-i}} a_i^{\frac{1}{k-i}} \\\\ a_j^{\frac{1}{k-j}} &=& a_k^{\frac{k-i-k+j}{(k-j)(k-i)}} a_i^{\frac{1}{k-i}} \\\\ a_j^{\frac{1}{k-j}} &=& a_k^{\frac{j-i}{(k-j)(k-i)}} a_i^{\frac{1}{k-i}} \\\\ \mathbf{a_j} &=& \mathbf{a_k^{\frac{j-i}{k-i}} a_i^{\frac{k-j}{k-i}} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_j} &=& \mathbf{a_k^{\frac{j-i}{k-i}} a_i^{\frac{k-j}{k-i}} } \\\\ && \boxed{a_{14} =375=a_i\quad i = 14\\ a_{2014}=a_j \quad j=2014\\ a_{4014}=135=a_k\quad k=4014} \\\\ a_{2014} &=& 135^{\frac{2014-14}{4014-14}} 375^{\frac{4014-2014}{4014-14}} \\\\ a_{2014} &=& 135^{\frac{2000}{4000}} 375^{\frac{2000}{4000}} \\\\ a_{2014} &=& 135^{\frac{1}{2}} 375^{\frac{1}{2}} \\\\ a_{2014} &=& \sqrt{135*375} \\ a_{2014} &=& \sqrt{50625} \\ \mathbf{a_{2014}} &=& \mathbf{225} \\ \hline \end{array}\)

The 2014th term is 225

As follows:

Here is the result for part a.  Use the same approach for part b.

Rearrange the first equation to get y = 10 - x.

Plug this into the second equation:  x^2 +(x - 10)^2 = 56

Expand this and rearrange to get a quadratic in x.  Solve for the two values of x. then substitute these back into y = 10 - x to find the corresponding two values of y.

The graph below shows that the value of k must lie between 0 and 100.  Clearly, you want the value of k that just clips the red graph at x = 50, so find the value of y on the red graph at x = 50 using the first equation, plug that into the second equation together with x = 50 and rearrange the resulting expression to find k.

Edit:  The value of x = 50 is incorrect - see below.

Draw \(\vec{a}\) anywhere on the coordinate plane. The origin seems like a convenient starting point

Draw \(-\vec{b}\), the opposite of \(\vec{b}\), and connect the tail of this vector to the head of \(\vec{a}\).

Draw \(\vec{a}-\vec{b}\), the resultant vector, by starting the tail of this vector at the tail of \(\vec{a}\) and by having the head connect to the head of \(\vec{b}\).

Now, you're done!

Let's define a few variables to start.

Let A = one digit of the three-digit number

Let B = one digit of the three-digit number

Let C = one digit of the three-digit number

For problems like these, it is generally helpful to find an expression that relates everything together.  \(ABC, ACB, BAC, BCA, CAB, CBA\) represent all the six different 3-digit numbers. We can then represent the sum of all the unique 3-digit numbers with the expression \(222(A+B+C)\) because each digit is added twice in the units, tens, and hundreds digit. 

Also, it is generally helpful to find a bound of some kind that restricts the possibilities. This way, it is not necessary to check every single possibility mindlessly. We can achieve this by recognizing that we can represent the sum in a different way. The sum must be at least \(3231+123=3354\) and must not exceed \(3231+987=4218\).

With this information, we can solve an inequality that restricts the possibilities for \(A+B+C\).

\(3354\leq 222(A+B+C)\leq4218\\ 15.108...\leq A+B+C \leq 19\)

Of course, in the context of this problem, \(A+B+C\) is restricted to the set of whole numbers. Therefore, \(A+B+C\) equals 16,17,18, or 19. After this point, we will just test all the possibilities since we only have 4 to check anyway.

We can represent the 6th number as \(222(A+B+C)-3231\). Below, I have included a table for each possibility for \(A+B+C\):

From this table, only one of possibilities for the 6th number reigns supreme, 765, because 765 is the only number where its digits add to its corresponding \(A+B+C\) value.

 "The probability that I choose an I is 1/9, and the probability that I pick another I right after is 1/8"

Let me get this straight: on the 1st draw, the probability (1/9) is lesser than on the second one (1/8) ?!?!

"x" is opposite the right angle    

Both triangles are 30-60-90

Let the  hypotenuse of a smaller triangle be a Y

Y = 1 / cos(60°) = 2

X = 2 / sin(30°) = 4   

Graphing  is definitely WAY easier  than an algebraic method

This shows  that  we really have the area of a simple region  (a rectangle)

https://www.desmos.com/calculator/xood3yofrf

One dimension of the rectangle  is   sqrt [( 1.5^2 + 1.5^2]  = 1.5sqrt (2)

And the other  is    sqrt [ 2.5^2 + 2.5^2]  = 2.5sqrt(2)

The area is   ( 1.5) (2.5) sqrt (2) * sqrt (2)  =  3.75 * 2  =    7