Answer:
9/2 if n goes to infinity and that the 2n^3 is under the whole expression
Step-by-step explanation:
Let me clear this .
find limit (9n^3 + 5*n - 2)/ (2n^3)
as n --> infinity
Did I put the parentheses in the right spot?
because if you leave it the way you did, then the whole expression goes to positive infinity as n goes to infinity But I will do this with parentheses
so
find limit (9n^3 + 5*n - 2)/ (2n^3)
simplify expression
limit (9/2) + 5/(2n^2) - 1/(n^3)
= (9/2) + 0 - 0
= (9/2)
9514 1404 393
Answer:
B, C
Step-by-step explanation:
The sums are ...
A) 59.45
B) 118.16
C) 18.97
D) 27.76
Sums B and C have 8 in the ones place.
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You can add the ones-place digits. If the resulting ones-digit is 7 or 8, then add the tenths-place digits. If the carry from that sum, added to the sum of ones-place digits, gives a value of 8 or 18, then you found a sum with a ones digit of 8.
The answer to what the length of the leg would be is 15.
You would do this problem by first writing down your Pythagorean Theorem, which is a^2 + b^2 = c^2.
Since we have our hypotenuse which is c^2 in our equation, we would write or insert the number we have.
So our equation could be that a or b leg equals 20, it doesn’t matter which one.
So we could write, 20^2 + b^2 = 25^2. So we don’t know what b leg is.
First we should figure out what 20^2 is and what 25^2 is.
20^2 is 400 and 25^2 is 625.
Our equation now comes to 400 + b^2 = 625.
Now we take 400 and subtract it from
625 -> 400 + b^2 = 625
-400.
So 625 - 400 comes out to be 225.
Lastly instead of squaring or putting 225 to the second power, we do the opposite.
So instead of squaring 225 we must square root 225. √ 225 .
The square root of √ 225 comes out to be 15.
Answer:
I cant see it?
Step-by-step explanation: