Math Is Fun Forum

Hi. My snail speed math progression has carried me to a bunch of word exercises in the precalculus book that i'm trying to go through for most of this year, and i am pitifully terrible at word problems:

So my attempt at solving this is
1) find height function (i don't know another way of finding this without making width and length equal):
        h = 1.62x
        x^2 * 1.62x = 140
        h = 140/1.62x^2
2) find surface area function and plug in the height function:
        2x^2 + 4xh
        2x^2 + 4x(140/1.62x^2)
        2x^2 + 560x/1.62x^2
3) I feel like i should find the slant of this function, then find the vertex of this slant, which will give me x, then i should plug that number in surface area function, and get the minimal surface area. Of course none of this is working out for me, so could anyone lend me a helping hand here?

I'm reading the "structure and interpretation of computer programs" book (that i probably should not touch) and i'm trying to understand an explanation for an exercise.

-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Provided we have n types of coins 1, 5,10 ,25 ,50 (so if n = 3, then the type-of-coin(n) = 10), how many ways can we make change of m dollars (in cents)?

for f(m, n):
if m = 0 -> 1
if m < 0 or n = 0 -> 0
else -> f( m, (n - 1) ) + f( (m - type-of-coin(n)), n)
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
So the problem is "What are the orders of growth of the space and number of steps used by f(m, n) as m increases? ". The answer in http://wiki.drewhess.com/wiki/SICP_exercise_1.14 says that "It's then straightforward to show that the number of steps required to evaluate f(m 5) is θ(m^5). "

I don't understand what θ(m^5) means here, because if i take 11 cents and put it into the function above, i get f(11, 5) = 4 in 55 steps. Now if i take 12 cents, i get that f(12, 5) = 4 in 63 steps. So how should i interpret θ(m^5) here, 11^5 is 161051, not 55?

http://wiki.drewhess.com/wiki/SICP_exercise_1.14
https://tobilehman.com/images/blogimg/cc_11_5.png
https://mitpress.mit.edu/sicp/full-text/sicp/book/node17.html

The exercise:

Use polynomial long division to perform the indicated division. Write the polynomial in the form p(x) = d(x)q(x) + r(x).

(2x^3-x+1) / (x^2+x+1)

The answer given by the book is:

(2x^3-x+1) = (x^2+x+1)(2x-2)+(-x+3)

What i don't understand is how am i supposed to find q(x) and r(x) when (x^2+x+1) cannot be factored into smaller factors (which i could use for the long division, to find the quotient and the remainder).

Did you always love math and if not, how did you come to love it?

I have a new found interest in math (used to hate it), but i feel like i still have to drag myself to meet my self set minimal quota of an hour per day (though i often get frustrated with some exercise and then spend the next 5 hours trying to understand it). I would love nothing more than to spend whole days sitting there, doing exercises, being fascinated with new concepts and ideas, but i feel like i get fatigued pretty fast. Looking at the 10000 hour rule, and how it's author came to that number, it seems like those who are the masters of their craft like nothing more than to constantly do that one thing, bordering on levels of autism. So that leaves me contemplating, if one does not reach anything near that level of enthusiasm, is it just an exercise in futility? Is there some mythical math cupid that could shoot an arrow up my math ars*?

Hello, i just can't get out of the absolute value blackhole lol. The book i'm reading gave me this exercise, but no previous examples how to solve an inequality with two absolute values, so this is what i tried:

|x -3| - |2x +1| < 0

1) if x > 3

      (x-3) - (2x +1) < 0
      x > -4

2) if x < 3 and x >= 0

    -(x-3) - (2x + 1) < 0
     x > 2/3

3) if < 0

   -(x-3) + (2x +1) < 0
    x < -4

The answer given in the book is (-∞, -4) U (2/3, ∞), so all the "if" clauses are false (or at least the first one is). Why? And i'm kind of ashamed to admit, but i'm still not sure of what is it that i'm searching with these absolute value inequalities.

Good day good people.

I have a function: f(x) = x^2
Its point:              (-2, 4)
Second function: g(x) = (x+2)^2 - 3
Its point:              (-4, 1)

So the second function is a transformation of the first function:

g(x) = (x+2)^2 - 3  =   f(x+2) - 3

I understand why -2 (x in point one) becomes -4 (x in point two):

x + 2 = -2
x = -4

Now i don't understand (conceptually) why 4 (y in point one) becomes 1 (y in point two). Since i have changed the input in the transformation function from x to (x+2), why doesn't the output reflect this change and i'm still left with 4, from which i then subtract 3? Shouldn't 4 become 16 (because i have -4 in my input now) from which i then subtract 3? I know the right sequence of actions here, i just don't understand why. I feel like i'm thinking about this in a completely wrong way.

Gooday.

I'm doing some exercises and i'm stuck at one:

|x| = 12 - x^2

I need to solve this equation. Well, when i try to solve this, i get two quadratic equations:

-1(x+4)(x-3) = 0

and

(x-4)(x+3) = 0

So the answers i get are:  x = -4, x = 4, x = 3, x = -3, however the book says only x = 3 and x = -3 are viable answers. I understand why this is by plugging in 3 and -3 in the first equations, however i don't understand how i should come to this conclusion computationally?