Equations, Inequalities & VICS, basic equations problem set 1
#12 a+b/b=21, what is a/b? How does the solution go down? (I thought contrary to the solution that you can not break fraction ie. a+b/b=21 to a/b +1=21, a/b=20) ?

Geometry

ch. 3 prblm # 7: why is volume of the super-giant can 8(pie,r^2,h) not 4 (pie,r^2,h)
ch. 3 prblm # 10 (lost)
Need help with Perpendicular Bisectors, specifically how to find the midpoint (ch. 5)[/b]

Hi alibakir,

Equations, Ineq, VICs: Ch. 1 In-Action Problem #12
You can break the fraction in this problem because the addition is within the numerator, and the b is thus the denominator of each part of the numerator. It might help to review each step:
(a + b)/b = 21
(a/b) + (b/b) = 21 (the book explanation kind-of skipped showing this step...)
(a/b) + 1 = 21
a/b = 20

You cannot break a fraction when there is addition or subtraction within the denominator. For example, b/(a + b) CANNOT be written as b/a + b/b. That's probably the situation you were thinking of...

Geometry: Ch. 3 In-Action Problem #7
The Giant can has radius=r and height=h. The Super-Giant can has radius=2r and height=2h.

Volume of a cylinder = (pi)(radius)^2(height)

Thus, the volume of the Super-Giant can = (pi)(2r)^2(2h) = (pi)(4r^2)(2h) = 8(pi)rh

Don't forget that when you square the radius of the Super-Giant can, you square the quantity 2r, which is 4r^2 (you have to square that 2 along with the r!)

Geometry: Ch. 3 In-Action Problem #10
A 12 inch by 10 inch by 8 inch cardboard box has 6 faces. However, opposite faces are identical, so there are 3 pairs of uniquely sized rectangular faces: the 12 by 10 face, the 10 by 8 face, and the 12 by 8 face.

This question asks "what is the largest possible volume of a right cylinder that can be placed inside the box?" A cylinder will sit on one inside surface of the box, with its radius limited by the size of that face, while its height will be limited by the 3rd dimension of the box:

If the can sits on the 12 by 10 face, maximum radius is 5 (=10/2) while maximum height is 8-->Volume = pi(25)(8) = 200pi
If the can sits on the 10 by 8 face, maximum radius is 4 (=8/2) while maximum height is 12.-->Volume = pi(16)(12) = 192pi
If the can sits on the 12 by 8 face, maximum radius is 4 (=8/2) while maximum height is 10.-->Volume = pi(16)(10) = 160pi

The largest possible can will have volume of 200pi, and will sit on the 12 by 10 face of the box, with a height of 8 inches.

Mid-point of a line segment
To find the midpoint of any line segment, you need to know the end points of the segment. The midpoint will always be the average of those end points. In other words, the x-coordinate for the midpoint is the average of the x-coordinates of the two end points. Similarly for the y-coordinate.

Hi, all

Please remember to post the FULL text of any problems (including answer choices, if applicable) when posting a question.

Not to be the forum police, but official policy is that the instructor who sees it will merely ask you to post the full text and not answer your question until you do. Other students will want to be able to follow along and if the instructors had to look up and post the text for all questions, we'd be able to handle maybe only a quarter of the volume of questions. Emily was having a slow day so she was nice enough to look up and post the questions, but in future, please post the full text of any problem about which you ask. :)