<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>
Pier pressure lolllllllllllllllllllllllll
Answer:
<em>The fraction of the beads that are red is</em>
Step-by-step explanation:
<u>Algebraic Expressions</u>
A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:
r:y = 2:3
y:b = 5:4
We are required to find r:s, where s is the total of beads in the bag, or
s = r + y + b
Thus, we need to calculate:
![\displaystyle \frac{r}{r+y+b} \qquad\qquad [1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7Br%2By%2Bb%7D%20%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B1%5D)
Knowing that:
![\displaystyle \frac{r}{y}=\frac{2}{3} \qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7By%7D%3D%5Cfrac%7B2%7D%7B3%7D%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B2%5D)
Multiplying the equations above:
Simplifying:
![\displaystyle \frac{r}{b}=\frac{5}{6} \qquad\qquad [3]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7Bb%7D%3D%5Cfrac%7B5%7D%7B6%7D%20%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B3%5D)
Dividing [1] by r:
Substituting from [2] and [3]:
Operating:
The fraction of the beads that are red is 
Answer:
<u>110.000</u>
<u></u>
Step-by-step explanation:
Selling at a loss of 5% means it was sold for 95% of its purchase value.
If 120.175 equals to 95% then 100%(the purchase value) is 126.500.
120.175/95*100 = 126.500 (plenty of other ways to calculate this, but this is one of the easiest)
So now we change to the dealer's perspective, he sold the car for 126.500 and for him that meant a 15% profit. If the purchase value is 100%, then the sale value is 115% (purchase value+15% profit)
Calculate the purchase value:
126.500/115*100 = 110.000
The dealer bought the car for 110.000, sold it for 126.500, so he made a profit of 16.500 (126.500-110.000) which is a 15% profit (15% of 110.000 is 16.500).
Answer:
The simplified expression is:
Step-by-step explanation:
We want to simplify:
We factor the numerator and the denominator to get:
Cancel out u-2 which is common to the numerator and the denominator yo get:
