Yo has to factor since we assume that
if you have xy=0, x and y=0
7n^2-48n+36=0
trial and error
(7n-6)(n-6)=0
set each to zero
7n-6=0
add 6
7n=6
divide 7
n=6/7
n-6=0
add 6
n=6
n=6/7 or 6
Answer:
24
Step-by-step explanation:
The question is saying, how many three digit numbers can be made from the digits 3, 4, 6, and 7 but there can't be two of the same digit in them. For example 346 fits the requirements, but 776 doesn't, because it has two 7s.
Okay, on to the problem:
We can do one digit at a time.
First digit:
There are 4 digits that we can choose from. (3, 4, 6, and 7)
Second digit:
No matter which digit we chose for the first digit, there is only going to be 3 of them left, because we already chose one, and you can't repeat that same digit. So there are 3 options.
Third digit:
Using the same logic, there are only 2 options left.
We have 4 choices for the first digit, 3 choices for the second, and 2 for the third.
Hence, this is 4 * 3 * 2 = 24 three-digit numbers that can be made.
Expanded Notation:
A. 654.362 = (6x100) + (5x10) + (4x1) + (3x0.1) + (6x0.01) + (2x0.001)
B. 125.384 = (1x100) + (2x10) + (5x1) + (3x0.1) + (4x0.01) + (8x0.001)
Answer:
B.3/2
Step-by-step explanation:
Answer:
9=x
Step-by-step explanation: