Solve for x; e^4=e^2x-1

4x and 6y

im not sure though

so, converting firstly the mixed fractions to improper fractions and then "adding".

answer: do we write it

step-by-step explanation:

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: