<span>The
content of any course depends on where you take it--- even two courses
with the title "real analysis" at different schools can cover different
material (or the same material, but at different levels of depth).
But yeah, generally speaking, "real analysis" and "advanced calculus"
are synonyms. Schools never offer courses with *both* names, and
whichever one they do offer, it is probably a class that covers the
subject matter of calculus, but in a way that emphasizes the logical
structure of the material (in particular, precise definitions and
proofs) over just doing calculation.
My impression is that "advanced calculus" is an "older" name for this
topic, and that "real analysis" is a somewhat "newer" name for the same
topic. At least, most textbooks currently written in this area seem to
have titles with "real analysis" in them, and titles including the
phrase "advanced calculus" are less common. (There are a number of
popular books with "advanced calculus" in the title, but all of the ones
I've seen or used are reprints/updates of books originally written
decades ago.)
There have been similar shifts in other course names. What is mostly
called "complex analysis" now in course titles and textbooks, used to be
called "function theory" (sometimes "analytic function theory" or
"complex function theory"), or "complex variables". You still see some
courses and textbooks with "variables" in the title, but like "advanced
calculus", it seems to be on the way out, and not on the way in. The
trend seems to be toward "complex analysis." hope it helps
</span>
Answer: "Count 2 and then count 31 more."
Step-by-step explanation:
We have the equation:
231 - 198
Now, the negative number is kinda ugly, so i will write it as:
200 - 2 = 198
Is a lot easier work with 200 and 2, than with 198.
Then the equation is now:
231 - (200 - 2)
And the left number we also have a "200", so it can be written as:
200 + 31 = 231
As this is a sum, we can ignore the parentheses.
200 + 31 - 200 + 2
31 + 2
Then the correct option is:
"Count 2 and then count 31 more."
Answer:
I assume you know Arithmetic Progression .
so, we have to find the first and last 4-digit number divisible by 5
first = 1000 , last = 9990
we have a formula, 
= a + (n-1)d
here, is the last 4-digit number divisible by 5.
n is the number of 4-digit even numbers divisible by 5
d is the common difference between the numbers, which is 10 in this case
a is the first 4-digit number divisible by 5
9990 = 1000 + (n-1)*10
899 = n-1
n = 900
Hence, there are 900 4-digit even numbers divisible by 5
Ask google or siri i dont see it cause it's sideways
