<span>If a=-2 and b=1, then
3a-2b
= 3(-2) -2(1)
= -6 -2
= -8</span>
How many 6-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, if repetitions of digits are allowed?
sveta [45]
There are 6 digits. Each digit can take ten different numbers except for the first digit since it cannot be zero.
So:
9 x 10 x 10 x 10 x 10 x 10
900000 numbers.
Another way of thinking about this is to just count up to 999,999. Obviously there are 999,999 different numbers here. But since our number has to have 6 digits in them, we have to delete 99,999 numbers. Thus there are 900,000 different numbers.
Answer:
a = -2 z^-3 c^6
Step-by-step explanation:
equations are given as
a = 2b^3 .....1 and
b = -1/ z c^-2 .....2
put the value of b in eq 1
a = 2(-1/z c^-2)^3
a = 2(-1/z^3 c^-6)
or a = -2 z^-3 c^6
Answer:
i--dk but may the 4th is "May the fourth be with you" and may the fifth is "Revenge of the fifth"
Step-by-step explanation:
x=angle CDB (vertical angles)
=angle ABE (corresponding angles, AB parallel to CD)
=132-90 (exterior angle 132=sum of angles ABE and AEB)
=42 degrees
