TheXSquaredFactor

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:

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

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.

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.

I did this same exercise, and I got 387!

If you are curious about the code I wrote, then here you go:

public static void main(String[] args)
    {
        int numPassed = 0;
        int numToTest = 0;
        double temp = 0.0;
        for (int hundredDigit = 0; hundredDigit<=9; hundredDigit++)
        {
            for (int tenDigit = 0; tenDigit<=9; tenDigit++)
            {
                for (int oneDigit = 0; oneDigit<=9; oneDigit++)
                {
                    numToTest = hundredDigit*100+tenDigit*10+oneDigit;
                    temp = Math.sqrt((double) numToTest);
                    if (temp != Math.floor(temp) && numToTest%2==1 && numToTest%5!=0)
                    {
                        System.out.println(numToTest);
                        numPassed++;
                    }
                }
            }
        }
        System.out.println(numPassed);
    }

Here is the list, if you are curious: 3,7,11,13,17,19,21,23,27,29,31,33,37,39,41,43,47,51,53,57,59,61,63,67,69,71,73,77,79,83,87,89,91,93,97,99,101,103,107,109,111,113,117,119,123,127,129,131,133,137,139,141,143,147,149,151,153,157,159,161,163,167,171,173,177,179,181,183,187,189,191,193,197,199,201,203,207,209,211,213,217,219,221,223,227,229,231,233,237,239,241,243,247,249,251,253,257,259,261,263,267,269,271,273,277,279,281,283,287,291,293,297,299,301,303,307,309,311,313,317,319,321,323,327,329,331,333,337,339,341,343,347,349,351,353,357,359,363,367,369,371,373,377,379,381,383,387,389,391,393,397,399,401,403,407,409,411,413,417,419,421,423,427,429,431,433,437,439,443,447,449,451,453,457,459,461,463,467,469,471,473,477,479,481,483,487,489,491,493,497,499,501,503,507,509,511,513,517,519,521,523,527,531,533,537,539,541,543,547,549,551,553,557,559,561,563,567,569,571,573,577,579,581,583,587,589,591,593,597,599,601,603,607,609,611,613,617,619,621,623,627,629,631,633,637,639,641,643,647,649,651,653,657,659,661,663,667,669,671,673,677,679,681,683,687,689,691,693,697,699,701,703,707,709,711,713,717,719,721,723,727,731,733,737,739,741,743,747,749,751,753,757,759,761,763,767,769,771,773,777,779,781,783,787,789,791,793,797,799,801,803,807,809,811,813,817,819,821,823,827,829,831,833,837,839,843,847,849,851,853,857,859,861,863,867,869,871,873,877,879,881,883,887,889,891,893,897,899,901,903,907,909,911,913,917,919,921,923,927,929,931,933,937,939,941,943,947,949,951,953,957,959,963,967,969,971,973,977,979,981,983,987,989,991,993,997,999

I assume the discrepancy occurred because Guest included 1, which should not be included because it is a square number.

It is vitally important to use a diagram, I would say. Luckily, I have taken the time to create one for you so that you can follow along. I have put all the information we know into the diagram below:

In order to find the perimeter, we need to express the length of each side of \(\triangle EFC\) in terms of x. \(\angle B\) is a right triangle, so by Pythagorean's theorem, \(AC=10\) and \(EC = 10-x\).

\(\triangle ACB \cong \triangle ECF \) by Angle-Angle Similarity Postulate, and we will use this similarity in order for us to express the remaining side lengths in terms of x like so:

\(\frac{10-x}{10}=\frac{EF}{8}\quad\quad\quad\frac{10-x}{x}=\frac{CF}{6}\\ EF = 8-\frac{4}{5}x\quad\quad CF=6-\frac{3}{5}x\) 

Now, find the perimeter by taking the sum of all the sidelngths of the desired triangle.

\(P=EC+EF+CF\\ P=(10-x)+(8-\frac{4}{5}x)+(6-\frac{3}{5}x)\\ P=24-\frac{12}{5}x\)

\((x-h)^2+(y-k)^2=r^2\) is the center-radius form for a circle such that \((h,k)\) is the coordinates of the center of the circle on the coordinate plane and \(r\) is the radius.

1) Find the equation of the circle centered at (1,-4) with radius 3

Since \((h,k)\)  of \((x-h)^2+(y-k)^2=r^2\) represents the coordinates of the center and \((1,-4)\) is already known to be the center, then we know that \(h=1\text{ and }k=-4\) . \(r\) is the radius, so r=3.

Substitute these values into the center-radius form of a circle to generate the equation of this particular circle. 

\(h=1,k=-4,r=3\\ (x-h)^2+(y-k)^2=r^2\\ (x-1)^2+(y-(-4))^2=3^2\\ (x-1)^2+(y+4)^2=9\)

2) I encourage you to use the information I gave you above to try and work out the second problem.

3) Find the distance between the points \((0,8)\text{ and }(2,-5)\).

We can use the distance formula, \(d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\) . Substitute the points that we want to find the distance from. 

I use the method StarStrike described above to post images, as well. 

Hey, DeathlyHallows!

It looks like you want to simplify \(\frac{x^2-5x+6}{\textcolor{red}{2x-1+\frac{x}{2}}}\) . Let's transform all the terms in the denominator such that they all have common denominators; this way, we will be able to add them together. 

\(\textcolor{red}{2x-1+\frac{x}{2}}\Rightarrow\textcolor{red}{\frac{4x}{2}-\frac{1}{2}+\frac{x}{2}}\)

Notice that I have not changed the value of the expression; I just created a common denominator. Now, combine like terms. 

\(\textcolor{red}{\frac{4x}{2}-\frac{1}{2}+\frac{x}{2}\\ \frac{5x}{2}-\frac{1}{2}\\ \frac{5x-1}{2}}\)

Therefore, \(\frac{x^2-5x+6}{2x-1+\frac{x}{2}}=\frac{x^2-5x+6}{\frac{5x-1}{2}}\) . Now, it is time to simplify. 

\(\frac{x^2-5x+6}{\frac{5x-1}{2}}*\frac{2}{2}\\ \frac{2(x^2-5x+6)}{5x-1}\\ \frac{2(x-2)(x-3)}{5x-1}\)

Even after I factored the numerator completely, I did not find any common factors present in both the numerator and the denominator, so this expression is already in simplest form. 

Quincy suspects that the two triangles above are similar, but there is not enough information to prove this. This is because \(\triangle JKL\) only has two defined side lengths, which is not enough to determine the other unknown angles or sides. 

2a) The SAS Similarity Theorem proves the similarity of two triangles by showing that the following is true:

• Both side lengths of one triangle are proportional to the side lengths of the other triangle.
• The included angle in one triangle is congruent to the corresponding angle of the other triangle.

Quincy would need the measure of \(\angle K\). This is the included angle of \(\overline{JK}\text{ and }\overline{KL}\) . We can determine the measure of the corresponding angle, \(\angle Q\) , by using the Law of Cosines. 

2b) The SSS Similarity Theorem proves the similarity of two triangles by showing that the following is true:

• All three side lengths of one triangle are proportional to the corresponding sides of another triangle.

Quincy would need the length of \(\overline{JL}\) . We already know the lengths of the other two side lengths of \(\triangle JKL\) and all the side lengths of \(\triangle PQR\), so this is the only extra bit of information needed to prove similarity. 

2c) The AA Postulate proves the similarity of two triangles by showing that the following is true:

• Two corresponding angles of the two triangles are congruent. 

Of course, Quincy would need to know the measure of two angles before proving their congruency, so it is an unlikely candidate to prove their similarity. Yes, it is possible to do, but it would be quite unnecessary.