$e\cdot e^x -e^{-2}=-2$
$\implies e^{x+1}=e^{-2}-2$
note that RHS is negative. (because e with negative exponent is less than 1)
and LHS is always positive.
so there cannot be any solution
Deijah has $6.25.
Now, this 6.25 is comprised of 0.05s and 0.25s.
We know that there are 12 0.25s.
We now want to know how many remaining 0.05s there are.
Again, we know that the number of 0.05s he has, which is 12, multiplied by 0.05, plus the number of 0.25s he has, multiplied by 0.25, equals 6.25.
Thus, the answer is A, 0.25 x 12 + 0.05 x n = 6.25.
Answer:
0
Step-by-step explanation:
A= (4-1)^3
simplify A to be 3^3
Which gives us 27
B=(2*3)^2-9
simplify B
first multiply the 2 numbers in paranthesis which gives us 6. raise it to the power of 2 which is 39 and then subtract 9. Gives us 27.
C=15^3*4-12
Simplify the exponent first. 3*4 gives us 12 and 12-12 equals 0. Anything raised to the power of 0=1
If A-B^C is the equation we can write 27-27 raised to the power of 1 which is 0
