tomtom

The answer is g(x) = -5x^2 + 7x + 4.

The answer is g(x) = -5x^2 + 7x + 4.

The answer is (sqrt(2) + 1)/3.

The answer is (sqrt(2) + 1)/3.

Here's how we can find the number of 3-digit numbers with exactly 9 factors:

Factorization: A number with 9 factors must be a perfect square of a number with 5 factors (since the factors come in pairs). For example, 256 has 9 factors because it's the square of 16, which has 5 factors (1, 2, 4, 8, 16).

Perfect Squares: To find 3-digit perfect squares, we can start by listing the perfect squares from 100 to 999: 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900.

Counting Factors: Now, we need to check which of these perfect squares have a number with 5 factors. We can do this by prime factorizing each number and counting the number of factors using the formula: (exponent of factor 1 + 1) * (exponent of factor 2 + 1) * ...

Here's a table showing the prime factorization and number of factors for each perfect square:

Perfect SquarePrime FactorizationNumber of Factors

100 2^2 * 5^29

121 11^23

144 2^4 * 3^225

225 3^2 * 5^29

256 2^89

324 2^2 * 3^415

400 2^4 * 5^225

441 3^2 * 7^29

484 2^2 * 11^29

529 23^23

676 2^2 * 13^29

841 29^23

900 2^2 * 3^2 * 5^227

Counting Numbers with 9 Factors: Finally, we count the number of perfect squares in the table that have exactly 9 factors. There are four such numbers: 100, 256, 441, and 841.

Therefore, there are four 3-digit numbers with exactly 9 factors.

We are asked to find the complex number \( z \) that satisfies the equation:

\[
(1+i)z - 2z^* = -11 + 25i
\]

where \( z^* \) denotes the conjugate of \( z \). Let \( z = x + yi \), where \( x \) and \( y \) are real numbers. Then, the conjugate \( z^* \) is given by:

\[
z^* = x - yi
\]

### Step 1: Substitute \( z = x + yi \) and \( z^* = x - yi \) into the equation

Substituting these into the equation, we get:

\[
(1+i)(x + yi) - 2(x - yi) = -11 + 25i
\]

### Step 2: Expand the terms

First, expand \( (1+i)(x + yi) \):

\[
(1+i)(x + yi) = 1 \cdot x + 1 \cdot yi + i \cdot x + i \cdot yi = x + yi + xi + i^2y = (x - y) + i(x + y)
\]

Next, expand \( -2(x - yi) \):

\[
-2(x - yi) = -2x + 2yi
\]

Substituting these into the equation:

\[
(x - y + 2y)i + (x - 2x - y) = -11 + 25i
\]

### Step 3: Combine like terms

Combining the real parts and imaginary parts, we have:

\[
(x - y - 2x) + (2y + x + 2yi) = -11 + 25i
\]

This simplifies to:

\[
-x - y + 2yi + (x - 2x) + (y + y)i = -11 + 25i
\]

### Step 4: Equate real and imaginary parts

Now, equate the real and imaginary parts of the equation:

For the real part:

\[
x - 2y = -11
\]

For the imaginary part:

\[
x + 2y = 25
\]

### Step 5: Solve the system of equations

We now solve the system of equations:

\[
x - 2y = -11
\]

\[
x + 2y = 25
\]

Add these two equations:

\[
(x - 2y) + (x + 2y) = -11 + 25
\]

This simplifies to:

\[
2x = 14
\]

So:

\[
x = 7
\]

Substitute \( x = 7 \) into one of the original equations:

\[
7 - 2y = -11
\]

Solve for \( y \):

\[
-2y = -18 \quad \Rightarrow \quad y = 9
\]

### Step 6: Write the complex number \( z \)

The complex number \( z \) is:

\[
z = x + yi = 7 + 9i
\]

Thus, the solution is:

\[
\boxed{7 + 9i}
\]

We can divide this problem into two cases:

Case 1: Each row has exactly one child from each family.

Choose one child from each family for the first row: there are 3 choices

.

Arrange the remaining two children in the first row (siblings can't be together): there are 2!=2 ways.

Arrange the second row similarly: 3⋅2=6​ ways total.

Case 2: One row has two children from the same family.

There are two subcases depending on the arrangement of siblings in the rows:

Subcase 2a: The first child in each row is from the same family.

Choose one pair of siblings: 3 choices

.

Arrange the siblings within the pair: 2 ways.

Arrange the remaining 4 children (2 from another pair and 2 from the third pair) in the second row: 4! ways.

Overcount: we've counted the arrangement as if sibling order matters within a pair (which it doesn't) twice (once for each sibling in the first row). So, we divide by 2!⋅2!.

Total: 2!⋅2!3⋅2⋅4!​=36 ways.

Subcase 2b: The first child in the first row is NOT a sibling of the first child in the second row.

Choose one pair to have their children occupy the third seats in each row: 3 choices.

Arrange the remaining 4 children in the first row: 4! ways.

Overcount: similar to subcase 2a, we divide by 2!⋅2! to account for sibling order not mattering.

Total: 2!⋅2!3⋅4!​=36 ways.

Since Cases 1 and 2 are mutually exclusive, to find the total number of arrangements, we simply add the number of arrangements from each case:

6 + 36 + 36 = 78​.

This gives us a final answer of 78.

Problem 2)

Since BD=AD=AC then triangle ABC is isosceles with AB=AC. Therefore, ∠ABC=∠ACB=x (for some angle measure x ).

Since D is on side BC so BD=AD, then line AD is an angle bisector of ∠A. Therefore, ∠BAD=∠CAD=2∠A​.

Since the angles in a triangle sum to 180∘, then in triangle ABC, ∠A+∠B+∠C=180∘. Substituting x for ∠B and ∠C, we get ∠A+x+x=180∘.

Solving for ∠A, we get ∠A=180∘−2x.

We are also given that ∠BAD=∠CAD=2∠A​. Substituting the expression for ∠A that we just found, we get 2180∘−2x​=∠BAD=∠CAD.

Because the sum of the angles in a triangle is 180∘, then in triangle ABD, ∠BAD+∠ABD+∠ADB=180∘.

Substituting 78∘ for ∠ABD (since BD=AD=AC and triangle ABC is isosceles), we get 2180∘−2x​+78∘+90∘=180∘.

Solving for x, we get x=31∘​.

Therefore, angle ACB = 31 degrees.

Problem 1)

Since ∠ABC=78∘ and ∠ACB=47∘, then ∠BAC=180∘−78∘−47∘=55∘.

Let D be the foot of the altitude from B to AC. Then ∠ABD=∠ABC=78∘.

Triangle ABD is a 30-60-90 triangle, so ∠ADB=60∘ and ∠BAD=30∘.

Since I is the incenter, ∠BAI=∠CAI=∠BAC/2=27.5∘. Then ∠ABI=∠BAI+∠BAD=57.5∘.

Triangle ABI is a 30-57.5-92.5$ triangle, so ∠BID = ∠ABI/2 = 46.25∘​.

Since a rotation about point O maps A to B, B to C, and C to D, we know the following:

Angle AOD: This is given as 24 degrees.

Full Rotation: A full rotation around a point is 360 degrees.

Possible Rotations:

There are three possibilities for the rotation that maps A to B, B to C, and C to D, resulting in three possible measures for angle AOB:

Case 1: Single Rotation of 24 Degrees

In this case, the rotation about O maps A to B by 24 degrees clockwise, B to C by 24 degrees clockwise, and C to D by 24 degrees

clockwise, resulting in a total rotation of A to D of 24 + 24 + 24 = 72 degrees.

Since the full rotation is 360 degrees, the remaining angle for AOB must be 360 - 72 = 288 degrees.

However, angle measures are typically represented between 0 and 180 degrees. We can achieve this by subtracting a multiple of 180 (full rotations) from 288:

AOB = 288° - 180 = 108.

Case 2: Double Rotation (360 + 24 Degrees)

Here, the rotation about O maps A to B by a full rotation (360 degrees) followed by a 24-degree clockwise rotation. This effectively brings B back to its original position and then continues the rotation to C and D.

The total rotation for AOD in this case becomes 360 + 24 + 24 = 408 degrees.

Similar to case 1, we can adjust this to fit the 0-180 degree range:

AOB = 408° - 360° = 48°

Case 3: Triple Rotation (2 * 360 + 24 Degrees)

This case involves two full rotations followed by a 24-degree clockwise rotation from A to B. Again, the first two rotations effectively bring B back to its original position.

The total rotation for AOD becomes 2 * 360 + 24 = 744 degrees.

Adjusting for the 0-180 degree range:

AOB = 744° - 2 * 360° = 124°

Summary:

Therefore, the three possible degree measures for angle AOB, considering rotations between 0 and 180 degrees, are:

108° (Case 1)

48° (Case 2)

124° (Case 3)