<span>In Babylonia, tunnels were used extensively for irrigation; and a brick-lined pedestrian passage some 3,000 feet (900 meters) long was built about 2180 to 2160 BC under the Euphrates River to connect the royal palace with the temple. Hopefully this answer helps
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Answer:
x = 0, -4 and -3.
Step-by-step explanation:
The given expression is "3x to the power of 3 + 21x squared + 36x = 0".
We can write this expression as follows :

Taking 3x common.

x = 3

Hence, the solution of the given quadratic equation are x = 0, -4 and -3.
You have a 1/3 chance each time you role
So 1/3 x 12 = 4 times that you would role a 5 or a 6
Degree is the value of the highest power of the variable, and hence in this case is 8
Answer:

Step-by-step explanation:
The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

We first compute the n-th derivative of
, note that

Now, if we compute the n-th derivative at 0 we get

and so the Maclaurin series for f(x)=ln(1+2x) is given by
