-13 (q + 4) = 7q + 16
I think to solve it you do this first:
-13q + -52 = 7q + 16
And then I think you do:
-6q + -36
And thats all I can remember sorry ):
Or
That -6q + -36 might be the answer I think (:
Hope I could help (:
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.
Idk cuz im stupid but cant you divide 9,998÷5?
I don’t know I am stuck on thisss
Please provide more information. You need to add the sides of the polygon to find the perimeter.
