*Please explain on how to do each of these!
1. Let (x) be defined for all positive integer values of x as the product of all even factors of 4x. For example, (3)=12x6x4x2=576. What is the value of (5)? Someone says that it is f(5)= 20x10x2x4=1600 I DO NOT GET HOW THEY ARRIVED THAT ANSWER!
2. 2.3 ^5 log(is raised at the same level as 5 meaning not above 5) base 2.3 ^1
The answer is 1 but I do not get how?
3. If p and q are two different prime numbers greater than 2, and n=pq, how many positive factors, including 1 and n, does n have?
Do not get how the answer is 4!!!!????
4. How do you put this in a TI-83 graphing calc?
A= 9 -3 1
1 1 1
16 4 1
Matrix A is 3x3
X= x
y
z
Matrix X is a 3x1
B= -7
-3
21
Matrix B is 3x1
• Find AX=B
1. To find the value of (5), which is the product of all even factors of 4x, we first need to determine the even factors of 4*(5). The even factors of 20 are 2, 4, 10, and 20. Now we can calculate the product:
(5) = 2 * 4 * 10 * 20 = 1600
So the value of (5) is indeed 1600, not 20x10x2x4.
2. The expression 2.3^5 log(base 2.3)^1 can be simplified using the laws of exponents and logarithms.
Since the base of the logarithm is the same as the base of the exponential term, we can rewrite the expression as:
2.3^(5 log(base 2.3)^1)
Now, according to the logarithmic property log(b^a) = a*log(b), we can bring the exponent of 5 down:
2.3^(5 * 1)
Simplifying further, we perform the multiplication:
2.3^5
Calculating this value, we have:
2.3^5 = 64.363
Therefore, the answer is approximately 64.363, not 1.
3. To find the number of positive factors of n (which is the product of two different prime numbers, p and q), we can consider that any positive factor of n will have the form p^a * q^b, where a and b represent the possible exponents of p and q, respectively.
Since p and q are different prime numbers greater than 2, their only positive factors are 1, p, q, and n = p*q. Therefore, there are only 4 positive factors, including 1 and n.
4. To solve the equation AX=B using a TI-83 graphing calculator, follow these steps:
1. Enter the matrix A into the calculator:
- Press [2nd] [X^-1] (MATRIX) to access the matrix menu.
- Navigate to "Edit" and press [Enter].
- Enter the values of the matrix A into a 3x3 matrix.
2. Enter the matrix B into the calculator:
- Press [2nd] [X^-1] (MATRIX) to access the matrix menu.
- Navigate to "Edit" and press [Enter].
- Enter the values of the matrix B into a 3x1 matrix.
3. Set up the equation AX = B:
- Press [2nd] [X^-1] (MATRIX) to access the matrix menu.
- Navigate to "Math" and then "B: Ax=B" and press [Enter].
- Select matrix A as the A matrix, matrix X as the X matrix, and matrix B as the B matrix.
- Press [Enter] to calculate the solution.
The calculator will solve for the values of x, y, and z in the equation AX = B and display the result in matrix X, which is a 3x1 matrix.