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        A_sector  =  (83 / 360) • A_circle    

        A_circle  =  π r²    

        60 π  =  (83 / 360) • π r²      For the next step, cancel out the π's      

                                                     and divide both sides by (83 / 360)   

        r²  =  60 • (360 / 83)  =  260.24        

        r  =  sqrt(260.24)  =  16.13   

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Since ∠APB=∠PBA, triangles ABP and PBA are isosceles. This means that ∠ABP=∠BAP=21​(180∘−∠APB).

Similarly, since ∠BPC=∠PBC, triangles BPC and PBC are isosceles. This means that ∠BPC=∠PBC=21​(180∘−∠BPC).

Summing the angle measures in △ABC gives $ \angle ABC + \angle ACB + \angle BAC = 180^\circ$. Since ∠APB=∠PBA and ∠BPC=∠PBC, we can rewrite this as [ \angle ABP + \angle BAP + \angle BPC + \angle PBC + \angle BAC = 180^\circ. ] Substituting what we found for each angle measure, we get [ \frac{1}{2} (180^\circ - \angle APB) + \frac{1}{2} (180^\circ - \angle APB) + \frac{1}{2} (180^\circ - \angle BPC) + \frac{1}{2} (180^\circ - \angle BPC) + \angle BAC = 180^\circ. ] Simplifying the left side gives [ 2 \cdot 180^\circ - (\angle APB + \angle BPC) + \angle BAC = 180^\circ. ] We are given that ∠ABC−∠ACB=24∘, which is the same as ∠BAP−∠BPC=24∘ since ∠APB=∠BAP and ∠BPC=∠PBC. Substituting again, we get [ 2 \cdot 180^\circ - (\angle BAP + \angle BPC) + \angle BAC = 180^\circ \Rightarrow 2 \cdot 180^\circ - (180^\circ + 24^\circ) + \angle BAC = 180^\circ. ] Solving for ∠BAC gives ∠BAC=36∘.

Since the angles in a triangle sum to 180∘, we can find ∠PBC as follows: [ \angle PBC = 180^\circ - \angle BPC - \angle BAC = 180^\circ - \frac{1}{2} (180^\circ - \angle BPC) - 36^\circ. ] Substituting ∠BAP−∠BPC=24∘ again, we get [ \angle PBC = 180^\circ - \frac{1}{2} (180^\circ - (\angle BAP + 24^\circ)) - 36^\circ = 180^\circ - 90^\circ - 12^\circ = \boxed{78^\circ}. ]

Prime factorize 200: 200=23⋅52

To get a product of 200, Catherine must roll exactly 3 twos and 2 fives. There are 6 choices for the first die roll, then 5 choices left for the second, and so on. However, this overcounts the order she rolls the dice in (e.g., 2-5-2-5-2-2 is counted the same as 5-2-2-5-2-2).

To account for this, we divide by the number of times we've overcounted. Since every roll matters (there are no duplicate numbers besides the twos), we simply divide by the number of permutations of 6 items (where order matters). There are 6!=720 ways to order 6 distinct items.

So, the total number of winning sequences is:

7206⋅5⋅4⋅3⋅2⋅1​=5​

We can solve this problem by considering different cases for how the siblings are arranged across the two rows.

Case 1: Splitting Each Pair

In this case, we have one child from each pair in each row. There are 3!=6 ways to arrange the children in each row (since sibling order within a row doesn't matter). However, we've overcounted since we haven't considered which sibling from each pair sits in which row. So, for each arrangement, we need to multiply by 23 to account for the two choices (sibling 1 or sibling 2) for each of the three pairs. This gives us 6⋅23=48 arrangements for this case.

Case 2: Keeping Pairs Together

Here, both siblings from one or two pairs sit in the same row. There are two sub-cases to consider:

Two Pairs Together: There are (23​)=3 ways to choose which two pairs will sit together (the third pair will be split). Once chosen, there are 2!=2 ways to arrange the two siblings within each pair that sits together, and again 2 ways to decide which row the combined pair sits in. Finally, there are 2!=2 ways to arrange the remaining split pair in the other row. This gives a total of 3⋅2⋅2⋅2=24 arrangements for this sub-case.

One Pair Together: There's only one way to choose which pair will sit together. Within that pair, there are 2! ways to arrange them. There are then 2 ways to decide which row the pair sits in, and 3!=6 ways to arrange the remaining two pairs (who must be split) in the other row. This gives a total of 1⋅2⋅2⋅6=24 arrangements for this sub-case.

Adding the sub-cases of Case 2, we get a total of 24+24=48 arrangements.

Total Arrangements

Summing the arrangements from both cases, we get a total of 48(Case1)+48(Case2)=96​ ways to seat the siblings.

There are 3 students in the math club, and 3 are in both clubs, so there are 3−3=0 students in just the math club.

The science club has twice as many students as the math club, so it has 2⋅3=6 students.

If we add the number of students in each club (without double counting the 3 who are in both), we get 0+6=6 students.

However, this double counts the 3 who are in both clubs. So, the total number of students who are in the math club or science club (or both) is 6−3=3​ students.

Analyzing the function:

This function defines f(x) as the square root of x minus a nested sequence of square roots that keep getting smaller. Intuitively, as x decreases, the nested square roots will also decrease, bringing f(x) closer to an integer.

Finding a bound:

For f(x) to be an integer, x - sqrt(x - sqrt(x - ...)) must be a perfect square. Let the innermost square root be y. We can rewrite the function as:

f(x) = sqrt(x - y)

Squaring both sides:

x - y = f(x)^2

Since we want the largest possible three-digit x, let's consider the smallest possible value f(x) can take. The smallest perfect square greater than 10 (a three-digit number) is 121. So, let's assume f(x) = 11.

Substituting:

x - y = 11^2 = 121

This tells us that x needs to be at least 121 more than the innermost square root (y) to be a perfect square.

Finding the largest three-digit x:

We know x must be greater than 121 + y. Since y is a square root and squares are always non-negative, the largest possible value of y for a three-digit x would be the square root of the largest three-digit perfect square, which is 9^2 = 81.

Therefore, x must be greater than 121 + 81 = 202.

Checking values:

We can now check the largest three-digit perfect squares less than 202:

196 (14^2) - When plugged into the function, it results in a non-integer value.

169 (13^2) - This value works! f(169) = 13, which is an integer.

Therefore, the largest three-digit value of x such that f(x) is an integer is x = 169.

Five times a number is divided by $2$ more than that number. If the result is $6,$ then what was the original number?  

5x / (x + 2)  =  6  

5x               =  6 • (x + 2)   

              5x  =  6x + 12   

             – x  =  12   

                 x  =  – 12    

_.

 PQ = 8, QR = 5, and PR = 10.

\(f_{PR}(x)=0\\ 5^2=8^2+10^2-2\cdot 8\cdot 10\cdot cos\ P\\ 160\cdot cos\ \ P=64+100-25\\ cos\ P=\frac{64+100-25}{160}\\ P=29.686° \)

\(f_{PX}(x)=tan\ (\frac{29.686}{2})\cdot x\\ f_{PX}(x)=0.265x\\ sin\ R:sin\ P=8:5\\ sin\ \ R= \dfrac{8\cdot sin\ P }{5}=\dfrac{8\cdot sin\ 29.686° }{5}\\ R=52.4105°\\ f_{RQ}(x)=tan\ (-R)\cdot (x-10)\\ f_{RQ}(x)=tan\ (-52,4105°)\cdot (x-10)\\ f_{RQ}(x)=-1,3\cdot (x-10)\\\)

\(f_{PX}(x)=f_{PQ}(x)\\ 0.265x=-1.3(x-10)\\ 0.265x=-1.3x+13\\ x=\dfrac{13}{1.565}\\ \color{blue}x_X=8.3067\\ \color{blue}y_X=2.2013\)

\(f_{XY}(x)=m(x-x_X)+y_X\\ m=tan\ (90°-R)=tan\ (90°-52.4105°)\\ m=0.7698\\ f_{XY}(x)=0.7698(x-8.3067)+2.2013\\ f_{XY}(x)=0.7698x-6.3946+2.2013=0\\ \color{blue}x_Y=5.4472\\ \color{blue}y_Y=0\)

\(L_{XY}=\sqrt{y_X^2+(x_X-x_Y)^2}=\sqrt{2.2013^2+(8.3067-5.4472)^2}\\ \color{blue}L_{XY}=3.6087\)

\(\color{blue}The\ length\ of\ \overline{XY}\ is\ 3.6807.\)

 !

The question can be solved by specifying PR=10 instead of PR=1.

 !

Since M is the midpoint of EF and N is the midpoint of EH, we have FM=ME=EN=NH. This means triangles FMP and HNP are similar.

Furthermore, since EFGH is a parallelogram, then ∠FPH=∠HPN (alternate interior angles).

Therefore, triangles FPH and HNP are also similar by Angle-Angle  Similarity (AA).

Since the triangles are similar, we have the following proportion:

NPPH​=HPFP​

Cross-multiplying gives us PH2=FP⋅NP. However, we want to find PQ/FH. Since FMP and HNP are similar, we also have:

FMPQ​=PHFP​

Cross-multiplying again gives us PQ⋅PH=FM⋅FP. Substituting the first equation we found (PH2=FP⋅NP) into this second equation, we get:

PQ⋅PH=FM⋅FP=FM⋅PH2

​=FM⋅PH

Dividing both sides by PH (which is nonzero since P is on line segment FH), we get PQ=FM. Finally, since FM=ME=21​EF, we have:

FHPQ​=FHFM​=2FHEF​=21​​

Find the intersection points (A and B):

We need to solve the system of equations for x and y:

y = (x - 4)/2 (equation of the line)

x^2 + y^2 = 6 (equation of the circle)

Substitute the first equation for y in the second equation: x^2 + ((x - 4)/2)^2 = 6

Solve for x: x^2 + (x^2 - 8x + 16)/4 = 6 4x^2 + x^2 - 8x + 16 = 24 (multiply both sides by 4 to eliminate the fraction) 5x^2 - 8x - 8 = 0 (x - 2)(5x + 4) = 0

Therefore, x = 2 or x = -4/5.

Find the corresponding y values for each x:

For x = 2: y = (2 - 4)/2 = -1

For x = -4/5: y = (-4/5 - 4)/2 = -14/5

Midpoint coordinates:

The midpoint of points A (2, -1) and B (-4/5, -14/5) is: Midpoint = ((2 - 4/5) / 2, (-1 - 14/5) / 2) Midpoint = (6/5, -9/5)

Therefore, the midpoint of AB is (6/5, -9/5).

Problem 2:

We can rewrite the equation in the desired form by multiplying both sides by the denominator of the original equation, (x+7).

Steps to solve:

Multiply both sides by (x+7): y * (x + 7) = (x - 3)

Substitute y with its definition: (x - 3)/(x + 7) * (x + 7) = (x - 3)

Simplify: x - 3 = (x - 3)

Now, we can see that the equation holds true for any value of x (as long as x ≠ -7 to avoid dividing by zero in the original equation). This means we can set the right side to any constant value (c) to maintain the truth of the equation.

However, to achieve the desired form (x + a)(y + b) = c, we should choose c = 0 (since the right side is already 0).

Therefore, a = 0, b = -7 (as y is replaced by (x - 3)/(x + 7)), and c = 0.

'

The answer in the form a, b, c is: 0, -7, 0

Problem 1:

The function simplifies to f(x) = 2(x + 5)/(x + 3).  You can then see that the function is not defined for x = -3.

We can analyze the system of quadratic equations to determine the number of real solutions for different values of c. Here's how:

Analyzing the Discriminant:

The discriminant of a quadratic equation determines the nature of its roots (solutions). It is denoted by the symbol b2−4ac. In this case, considering the first equation (y = 6x^2 - 9x + c):

a = 6

b = -9

c (variable)

The discriminant (d) for the first equation is:

d = (-9)^2 - 4 * 6 * c

The number of real solutions depends on the value of the discriminant:

d > 0: Two real and distinct solutions (roots)

d = 0: One repeated real solution (root)

d < 0: No real solutions (complex roots)

Relating Discriminant to c:

We want to find the values of c that correspond to each case.

(a) Exactly one real solution:

For exactly one real solution (repeated root), the discriminant needs to be zero.

Therefore, we need to solve:

0 = (-9)^2 - 4 * 6 * c

This simplifies to:

c = \frac{81}{24} = \dfrac{7}{2}

(b) More than one real solution:

For more than one real solution (distinct roots), the discriminant needs to be positive.

Therefore, we need to solve:

0 < (-9)^2 - 4 * 6 * c

This simplifies to:

c < \dfrac{81}{24} = \dfrac{7}{2}

(c) No real solutions:

For no real solutions (complex roots), the discriminant needs to be negative.

Therefore, we need to solve:

0 > (-9)^2 - 4 * 6 * c

This simplifies to:

c > \dfrac{81}{24} = \dfrac{7}{2}

Summary:

(a) Exactly one real solution: c = dfrac{7}{2}

(b) More than one real solution: c < dfrac{7}{2}

(c) No real solutions: c > dfrac{7}{2}

Unfortunately, with the given information (side lengths PQ=8, QR=5, and PR=1), it is impossible to determine the length of XY. Here's why:

Triangle Inequality Violation: The side lengths provided violate the triangle inequality. The triangle inequality states that the sum of any two side lengths in a triangle must be greater than the third side length. In this case, PQ+PR=8+1=9, which is not greater than QR=5. Therefore, a triangle with these side lengths cannot exist.

Ambiguous Angle Bisector: Even if the side lengths were valid, the information provided wouldn't be sufficient to determine the length of XY. The angle bisector of angle P divides angle P into two ratios, which depend on the original measure of angle P. Without knowing the angle measure or additional information about the triangle (like area), it's impossible to determine the specific location of point X on side QR​. Consequently, the distance XY cannot be determined.

Since P, Q, and R are the midpoints of segments in triangle MNO, then triangle PQR is similar to triangle MNO with a scale factor of 21​. Therefore, the area of triangle PQR is (21​)2 times the area of triangle MNO, or 41​ the area of triangle MNO. We are given that the area of triangle PQR is 10, so the area of triangle MNO is 10⋅4=40.

Similarly, triangle JQR is similar to triangle MNO with a scale factor of 21​, so the area of triangle JQR is (21​)2 times the area of triangle MNO, or 41​ the area of triangle MNO. We know the area of MNO is 40, so the area of triangle JQR is 41​⋅40=10​.

Since ∠TSU=60∘ and ∠STU=45∘, then ∠SUT=75∘. Triangle STU is a 45-45-90 triangle, so SU=UT=2​⋅ST. Also, since M and N are midpoints, then SM=MT=NU=UT=2​⋅ST and TN=NS=US=2​⋅ST.

Triangle UMN is isosceles with UM=MN because they are both half of sides of an isosceles triangle. Similarly, triangle PNM is isosceles with PN=MN. Since ∠SUT=75∘, then ∠SUM=∠SUN=(180∘−75∘)/2=52.5∘. Likewise, ∠PNM=∠PMN=52.5∘.

Looking at triangle ZNM, we see that ∠ZNM+∠ZMN+∠MZN=180∘. Since ∠MZN is a right angle, then ∠ZNM+∠ZMN=90∘. Similarly, for triangle PNM, we have ∠PNM+∠PMN=90∘.

Adding all the angle measures together, we get the answer: $ \angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM = 90^\circ + 90^\circ + 52.5^\circ + 52.5^\circ + \angle NZM + \angle NPM. $

Since ∠NZM and ∠NPM are supplementary angles (they form a straight line), then ∠NZM+∠NPM=180∘. Therefore, the final answer is $ 90^\circ + 90^\circ + 52.5^\circ + 52.5^\circ + 180^\circ = \boxed{465^\circ}$.

We can solve for a and b by utilizing the given information about A and its effect on x and y, along with the property of matrix multiplication.

Here's how we proceed:

Analyze the equation (A^5)x = ax + by:

This equation states that applying matrix A to vector x five times consecutively (A raised to the power of 5) results in a linear combination of x and y, with coefficients a and b.

Utilize the information about A:

We are given that Ax = y and Ay = x + 2y. These equations define how A transforms x and y.

Express (A^5)x in terms of x and y:

We can't directly expand (A^5) as it's a high power. However, we can use the given information about A iteratively.

Start with the first equation: (A^2)x = A(Ax) = A(y).

Substitute from the first given equation: (A^2)x = Ay.

Use the second given equation: (A^2)x = x + 2y.

Similarly, we can continue:

(A^3)x = A((A^2)x) = A(x + 2y) = Ax + 2Ay (using the definition of matrix multiplication).

Substitute from the first given equation: (A^3)x = y + 2(x + 2y) = 3x + 4y.

We can repeat this process further, but the pattern should be clear.

Find an expression for (A^4)x and (A^5)x:

Following the established pattern, we can see that:

(A^4)x = 3(A^3)x = 3(3x + 4y) = 9x + 12y.

(A^5)x = 3(A^4)x = 3(9x + 12y) = 27x + 36y.

Substitute (A^5)x in the original equation:

The original equation is: (A^5)x = ax + by.

Substitute the expression we found for (A^5)x: 27x + 36y = ax + by.

Solve for a and b:

We want to isolate a and b. Since x and y are not multiples of each other, we can treat them as independent variables.

If we set y = 0, the equation becomes 27x = ax, which implies a = 27.

If we set x = 0, the equation becomes 36y = by, which implies b = 36.

Therefore, in the equation (A^5)x = ax + by, the values of a and b are:

a = 27

b = 36

Since O is the origin (0,0), the coordinates of P are of the form (a,4a) for some positive value a. Similarly, the coordinates of Q are of the form (b,5b) for some positive value b. We are given that OP=OQ, so a2+(4a)2​=b2+(5b)2​, which simplifies to a=b.

Therefore, the coordinates of Q are (a,5a), which means the slope of PQ​ is [\frac{(5a) - (4a)}{a - (0)} = \boxed{1}.]

We can use the given information about A and B acting on x and y to find the expressions for (AB)x and (BA)x.

Finding (AB)x:

We know (AB)x = B(Ax) since matrix multiplication is associative.

From the givens, Ax = y. Substitute: (AB)x = B(y).

We are also given By = 2y. Substitute: (AB)x = B(2y) = 2(By) = 2(2y) = 4y. Therefore, we can write (AB)x = 4y. This translates to a = 0 and b = 4.

Finding (BA)x:

Similar to (AB)x, we know (BA)x = A(Bx).

From the givens, Bx = x + y. Substitute: (BA)x = A(x + y).

We are also given Ay = x + 2y. Substitute: (BA)x = A(x + y) = Ay - 2y = (x + 2y) - 2y = x. Therefore, we can write (BA)x = x. This translates to c = 1

and d = 0.

In conclusion, the scalars that satisfy the equations are:

a = 0

b = 4

c = 1

d = 0

This means:

(AB)x = 4y

(BA)x = x