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Re-arrange by simplifying to :  

x^2 +12x       +y^2   +2y    =   10        Now complete the squares for x and y 

x^2 + 12x +36    +   Y^2  +2y +1    = 10  +36 + 1 

(x+6)^2       + ( y+1)^2  =   47             47 is r^2   For this circle 

         Area of a circle = pi r^2   = pi * 47 = 47*pi = 147.65  units^2 

Arrange into standard circle equation form by completing the squares for x and y 

x^2 - 10x    + 25       + y^2 + 10y    + 25   = 25 +25 -c

(x-5)^2                   + ( y+5)^2     = 50 - c            50 - c = radius ^2 

                                                                             50 - c = 1^2

                                                                                -c = -49      So    C = 49  

Finding the Measure of Minor Arc TU

Understanding the Problem

We have a circle with center O.

Points T and U lie on the circle.

Point P is outside the circle with PT and PU tangent to the circle.

Angle TPO = 33 degrees.

We need to find the measure of minor arc TU.

Solution

Key Concept: The angle formed by two tangents to a circle from an external point is equal to the difference between the intercepted arcs.

Steps:

Find angle TPU:

Since PT and PU are tangents to the circle, angle PTO and angle PUO are right angles (90 degrees each).

In triangle TPU, the sum of angles is 180 degrees.

So, angle TPU = 180 - angle PTO - angle PUO = 180 - 90 - 90 = 0 degrees.

This means that points P, T, and U are collinear.

Find angle TOP:

Since triangle OTP is a right triangle (OT is a radius perpendicular to tangent PT), angle TOP = 90 - angle TPO = 90 - 33 = 57 degrees.

Find the measure of minor arc TU:

Angle TOP is the central angle subtended by minor arc TU.

The measure of a central angle is equal to the measure of its intercepted arc.

Therefore, the measure of minor arc TU is 57 degrees.

So, the measure of minor arc TU is 57 degrees.

Given points \( A = (4, -4) \), \( B = (3, 8) \), and \( C = (-1, 2) \), and a point \( Q \) such that for any point \( P \) in the plane, the following equation holds:

\[
PA^2 + PB^2 + PC^2 = 3PQ^2 + k
\]

We need to find the constant \( k \).

### Step 1: Write down the expression for the sum of squares of distances

Let \( P = (x, y) \) and \( Q = (x_Q, y_Q) \). The squared distances from \( P \) to the points \( A \), \( B \), and \( C \) are:

\[
PA^2 = (x - 4)^2 + (y + 4)^2
\]

\[
PB^2 = (x - 3)^2 + (y - 8)^2
\]

\[
PC^2 = (x + 1)^2 + (y - 2)^2
\]

The sum of these squared distances is:

\[
PA^2 + PB^2 + PC^2 = \left[(x - 4)^2 + (y + 4)^2\right] + \left[(x - 3)^2 + (y - 8)^2\right] + \left[(x + 1)^2 + (y - 2)^2\right]
\]

Expanding these expressions:

\[
PA^2 = (x^2 - 8x + 16) + (y^2 + 8y + 16)
\]

\[
PB^2 = (x^2 - 6x + 9) + (y^2 - 16y + 64)
\]

\[
PC^2 = (x^2 + 2x + 1) + (y^2 - 4y + 4)
\]

Adding them together:

\[
PA^2 + PB^2 + PC^2 = \left[3x^2 + (-8x - 6x + 2x) + (16 + 9 + 1)\right] + \left[3y^2 + (8y - 16y - 4y) + (16 + 64 + 4)\right]
\]

Simplifying:

\[
PA^2 + PB^2 + PC^2 = 3x^2 - 12x + 26 + 3y^2 - 12y + 84
\]

Thus:

\[
PA^2 + PB^2 + PC^2 = 3(x^2 - 4x + \frac{26}{3}) + 3(y^2 - 4y + 28)
\]

### Step 2: Express the right-hand side

The expression for \( 3PQ^2 \) is:

\[
3PQ^2 = 3\left[(x - x_Q)^2 + (y - y_Q)^2\right] = 3\left[(x^2 - 2xx_Q + x_Q^2) + (y^2 - 2yy_Q + y_Q^2)\right]
\]

Expanding:

\[
3PQ^2 = 3x^2 - 6xx_Q + 3x_Q^2 + 3y^2 - 6yy_Q + 3y_Q^2
\]

### Step 3: Set up the equation

We equate \( PA^2 + PB^2 + PC^2 \) with \( 3PQ^2 + k \):

\[
3x^2 - 12x + 110 + 3y^2 - 12y + 84 = 3x^2 - 6xx_Q + 3x_Q^2 + 3y^2 - 6yy_Q + 3y_Q^2 + k
\]

By comparing coefficients, we get:

\[
-12x = -6xx_Q \quad \text{and} \quad -12y = -6yy_Q
\]

So:

\[
x_Q = 2, \quad y_Q = 2
\]

Now, matching the constant terms:

\[
110 + 84 = 3x_Q^2 + 3y_Q^2 + k
\]

\[
194 = 3(2^2) + 3(2^2) + k = 12 + 12 + k = 24 + k
\]

Thus:

\[
k = 194 - 24 = 170
\]

### Final Answer:

The constant \( k \) is \( \boxed{170} \).

the ratio of the area of the sphere to the area of the cone is 9/5 .

Finding |u+2w|

Understanding the Problem

We are given three pieces of information about complex numbers u and w:

|u| = 5

|w| = 3

|u+w| = 6

We need to find the value of |u+2w|.

Solution Approach

We will use the Law of Cosines for complex numbers to solve this problem.

Applying the Law of Cosines

Let θ be the angle between u and w. Then, we have:

|u+w|^2 = |u|^2 + |w|^2 + 2|u||w|cosθ

Substituting the given values:

6^2 = 5^2 + 3^2 + 253*cosθ

36 = 34 + 30cosθ

cosθ = 1/15

Now, consider |u+2w|^2:

|u+2w|^2 = |u|^2 + (2|w|)^2 + 2|u|(2|w|)cosθ

|u+2w|^2 = 5^2 + (23)^2 + 25*(23)(1/15)

|u+2w|^2 = 25 + 36 + 12

|u+2w|^2 = 73

Therefore, |u+2w| = √73.

So, the value of |u+2w| is √73.

To find the probability that all three pieces of a stick, which is 6 units long, are shorter than 6 units after it is broken at two random points, we can use a geometric interpretation.

1. **Setting Up the Problem:**

- Let the stick start at 0 and end at 6 (it has a total length of \( L = 6 \) units).

- Denote the two break points as \( X_1 \) and \( X_2 \), where \( 0 < X_1 < X_2 < 6 \).

- The resulting pieces of the stick will have lengths:

- Piece 1: from 0 to \( X_1 \) (length \( X_1 \))

- Piece 2: from \( X_1 \) to \( X_2 \) (length \( X_2 - X_1 \))

- Piece 3: from \( X_2 \) to 6 (length \( 6 - X_2 \))

2. **Condition for the Pieces:**

- We need each piece to be less than 6 units in length:

- \( X_1 < 6 \)

- \( X_2 - X_1 < 6 \)

- \( 6 - X_2 < 6 \)

- The inequalities simplify to:

- \( X_1 < 6 \) (which will always be true since \( X_1 < X_2 < 6 \))

- \( X_2 < 6 + X_1 \) (or simply \( X_2 < 6 \) due to initial constraints)

- \( X_2 > 0 \) (will also hold since \( X_2 > X_1 > 0 \))

3. **Valid Conditions:**

- The only actual restriction is that \( X_2 < 6 \) (which is always satisfied because \( 0 < X_1 < X_2 < 6 \)).

- The lengths of the pieces will always be less than the length of the original stick (6 units).

4. **Geometric Approach:**

- We represent the breaking points \( (X_1, X_2) \) in the coordinate plane (in the region where \( 0 < X_1 < X_2 < 6 \)).

- The total area of the triangle formed by these restrictions in this coordinate system is represented by:

\[
\text{Area of valid region} = \frac{1}{2} \times 6 \times 6 = 18
\]

- However, the actual region that must be considered is bounded by the triangle formed by \( (0,0), (6,0), (6,6) \). The area of the square is \( 6 \times 6 = 36 \).

5. **Calculating the Probability:**

- The area of the successful outcomes where all pieces are shorter than 6 units is the same as the area where the inequalities hold. As we analyzed, if we consider the constraints, the condition is satisfied for all relative placements of the breaking points.

- Thus the probability \( P \) of obtaining such a division where each piece is shorter than 6 units is simply:
\[
P = \frac{\text{area of valid choices}}{\text{total area}} = \frac{18}{36} = \frac{1}{2}
\]

Thus, the probability that all three resulting pieces are shorter than 6 units is \(\frac{1}{2}\) or 50%.

Looks like we got the same answers! AUnverifiedTaxPayer!

Nice work! :)

Also, I accidentally labeled #4 as #3. My bad. 

Alright, here are the solutions that I got. 

1. It is given that the two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, hundreds digit and units digit must be either 2 or 3 i.e. 2_2, 2_3, 3_2 or 3_3 which means that there are four possible numbers - 242, 263, 362 and 393.

Now, it is also given that - the difference between its reverse and it is 99. So 263 and 362 satisfy this condition. Hence, the sum of the three digits is 11 in each case.

2. It is given that the man has 40 litres container of milk. Also, he will drink 1 litre on the first day and refill the container with water, will drink 2 litres on the second day and refill the container, will drink 3 litres on the third day and refill the container, and so on till 40th day. Thus at the end of 40 days, he must have drunk (1 + 2 + 3 + 4 + ..... +38 + 39 + 40) = 820 litres of liquid.

Out of those 820 litres, 40 litres is the milk which he had initially. Hence, he must have drunk 780 litres of water.

3. As there are 11 balls along one side, it means that there are 11 layers of balls. The top most layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55 and 66 balls in the remaining layers.

Hence, the total number of balls is = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 = 286 balls

4. Let's start with JKL = 9 * LQ. Note that L appear on both the side. Also, after multiplying LQ by 9 the answer should have L at the unit's place. The possible values of LQ are 19, 28, 37, 46, 55, 64, 73, 82 and 91; out of which only 64, 73 and 82 satisfies the condition. (As all alphabets should represent different digits)

Now, consider PQR = 6 * LQ. Out of three short-listed values, only 73 satisfies the equation.

Also, ZYX = 3 * LQ is satisfied by 73.

Hence, Z=2, Y=1, X=9, P=4, Q=3, R=8, J=6, K=5, L=7

219/3 = 438/6 = 657/9 = 73

5. The old man temporarily added his dog to the 17, making a total of 18 dogs.

First son got 1/2 of it = 9

Second son got 1/3 of it = 6

Third son got 1/9 of it = 2 for a total of 17.

He then steals his dog back and goes away......

5. I THINK this is the solution. Unless,umm..idk

The old man temporarily added his dog to the 17, making a total of 18 dogs.

First son got 1/2 of 18 = \(1/2\cdot 18 = 9\)

Second son got 1/3 of 18 = \(1/3 \cdot 18=6\)

Third son got 1/9 of 18  = \(1/9 \cdot 18 = 2\) for a total of 17.

Then I think the old man stole his dog back...so yay?

Nice questions! I might not answer them every week, but it sure was a nice little brain teaser!

Thanks! ::)

3. First, let's note that If any number LQ is multiplied by 9, right most digit of the product will be (10 - Q).

So, from JKL/9 = LQ, we get, L = 10 - Q

LQ can be mathematically represented as\(10L + Q = 10(10 - Q) + Q = 100 - 9Q \)

Since \(PQR/6 = LQ\), \(PQR = 6LQ = 6(100 - 9Q) = 600 - 54Q\)

We now try various values for Q such as 1, 2, 3, .... , etc, till the middle digit of the product turns out to be the same as the value of Q used. We get the correct answer for Q at the very third attempt, namely 3. Hence,

\(L= 10 - 3 = 7\)

So we finally have

\(219/3 = 73 \\ 438/6 = 73 \\ 657/9 = 73\)

Thanks! :)

I will give you that you got the three right! Nice job, and the method for number 3 was my strategy too. Great job NotThatSmart!

3. My logic for this problem might be flawed, but let me give it my best shot. 

If One side of the bottom layer of a triangular pyramid has 11 balls, 

The first layer has 1 ball

The Second layer will have \(3 = (1 + 2) \)balls

The Third layer will have \(6=(1 + 2 + 3)\) balls.

The Fourth layer has\( 10=(1 + 2 + 3 + 4)\) balls.

The Fifth layer will have \(15=(1 + 2 + 3 + 4 + 5)\) balls

The sixth layer there are\( 21=(1 + 2 + 3 + 4 + 5 + 6)\) balls

So there will be \(28,36,45,55 \quad \text{and} \quad 66 \) balls in the remaining layers.

So, adding the total number of balls up, we get

\(1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 = 286 balls.\)

So 286 is the answer. 

Again, not sure about this one, should be right. 

Thanks! :)

2. Note that if Harry Puttar (nice name LOL) refills 2 litres and drinks 3 litres, he will drink 1 litre of milk every single day. 

Thus, he should run out of milk by the 40th day. 

The number of litres he drank can be calculated with the sequence

\(1+2+3+4+5+6+...+39+40 = 820\)

Of these total 820 litres, there WAS 40 litres of milk,  meaning he drank

\(820-40 = 780\)litres of water. 

So 780 is our answer. 

Thanks! :)

1. Alright, let's give this a shot. First off, there are 4 prime numbers between 1 and 10. 

We have \(2,3,5,7\)

However, since the product of these two prime numbers must be less than 10 (single digit number), the only two numbers that qualify is \(2 \cdot 3 = 6\)

So now, we have two different numbers to work with. \(263, 362\)

Since their the reverses, take the larger number, 362 is the original. 

We find that \(362-263=99\), meaning we found our number. 

Adding the digits up, we get

\(2+6+3 = 11\)

So 11 is our answer. 

Thanks! :)

It's the same number of painters, all painting at the    

same rate that they painted the 2000 sqft wall.  

The only thing that's different is that the new wall is    

3 times as big, so it will take 3 times as long to paint.   

\(3 \cdot 40 =  120 \) minutes to paint

Thanks! :)

.

WAIT A SECOND! Let's note something real quick. 

First, let's look at the first equation. We want only two equations, we isolate p in terms of r. 

We get that

\(p=3+2q\)

Subsituting this value of p into the second and third equation, we get the system with r and q, we get

\(q-2r=-2+q\\ 3+2q+r=9+3+2q\)

Now, focusing on the first equation of the new system, isolating r, we find that

\(r=1\)

Plugging in this value for r into the second equation, we have that

\(3+2q+1=2q+12\\2q+4=2q+12\\-8=0\)

THIS IS DEFINITELY NOT TRUE. 

In conclusion, we know that this system has NO SOLUTIONS. 

I think this is right...

Thanks! :)

Let's set up a system of equations to solve this problem. 

Let's let b be bagels

Let's let t be toast

And let's let m be muffins. 

From the problem, we have the system

\(2T + 1B = 3.15 \\ 1M + 1B = 3.50 \\ 1T + 2B + 3M = 8.15\)

Now, we want everything to be in terms of bagels. 

From the first two equations, we get the equations

\([3.15 - B ] / 2 = T \\ M = 3.50 - B\)

Now, plugging these values into the third equation, we get

\([ 3.15 - B ] / 2 + 2B + 3 [ 3.50 - B ] = 8.15 \\ [3.15 - B ] + 4B + 6 [ 3.50 -B ] = 16.30 \\ -3B + 3.15 + 21 = 16.30 \\ -3B = 16.30 - 3.15 - 21 \\ -3B = -7.85 \\ B = 7.85 / 3 ≈ $2.62\)

This is 262 cents or about 261 and 2/3 cents. 

Thanks! :)

First, let's simplify the equation a bit. We get

\(x^2 - 16x + y^2 + 22y = 86\)

Completing the square for BOTH x and y, we get the equation

\(x^2 - 16x + 64 + y^2 + 22y + 121 = 86 + 64 + 121 \\ (x - 8)^2 + (y + 11)^2 = 271\)

This is a  circle centered at (8, - 11) with r^2  = 217

Now, we can find the area. We simply have

\(Area = pi * r^2 = 271 pi ≈ 851.4\)

Thanks! :)