Last edited on 14 September 2016 - 03:29 AM by RegalRat

I need help, everyone! Please, assist me, and you will be rewarded with kind words of praise.


Math Homework Name: Bowser

1.) Solve for x: ax+b = cx+d.

2.) Solve the system of equations for the ordered pair (x,y):

3x+4y = M
5x+6y = N.

3.) A circle has area A. A square has area B. The perimeter P of the square is the same as the diameter of the given circle.

a.) What is A/B?

b.) Evaluate the sum S = 1 + (B/A) + (B/A)^2 + (B/A)^3 + … .

4.) Let quadratic polynomials p(x) = ax^2+bx+c and q(x) = dx^2+ex+f be given (each constant is non-zero).

a.) What is the sum of the roots of the fourth-order polynomial p(x)q(x) terms of the given constants?

b.) Suppose 3 is a zero of both p(x) and q(x). Give the four zeros of p(x)q(x).

5.) Consider the function f(x) = x ln(x^2).

a.) What is the average value of this function over an arbitrary positive interval [a,b]?

b.) What is the behavior of f(x) as x approaches zero?

c.) What's the equation of the tangent line to the graph of f(x) as x = 5?

These are just the warm-up problems! Help!

Thanks!

Posted on 14 September 2016 - 10:18 AM

OK, guys. After working hours upon hours, I think I got the first one at least.

Here's my work:

ax+b = cx+d

(a-c)x = d-b

x=(d-b)/(a-c).

I still need help on the others!

Posted on 14 September 2016 - 10:46 AM

Try using photomath, I use when I am confused about the question. Dunno that photomath can solve those ones though.

Posted on 14 September 2016 - 12:18 PM

Wow, Photomath is wacky. That's too easy!

Posted on 14 September 2016 - 02:25 PM

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Posted on 14 September 2016 - 02:29 PM

No, 8. Can you help me?

Last edited on 14 September 2016 - 03:16 PM by *deleted-49709

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Posted on 14 September 2016 - 03:26 PM

I am advanced.

Last edited on 14 September 2016 - 03:28 PM by *deleted-49709

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Last edited on 14 September 2016 - 03:28 PM by RegalRat

Please, post solutions on this thread.

Posted on 14 September 2016 - 03:29 PM

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Posted on 14 September 2016 - 03:30 PM

Tsk, tsk. This is why you aren't up to par with the high marks kids in your class.

Last edited on 14 September 2016 - 10:42 PM by *deleted-49709

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Posted on 14 September 2016 - 05:14 PM

Thank you for the help, Tanthegreat.

With the help of the Badlion community, I will now be able to do my math homework.

I will post solutions now.

1.) I did it in a previous post. Here it is again for completeness:

ax+b = cx+d

(a-c)x = d-b

x=(d-b)/(a-c).
——————————————————————
2.) Solve by elimination: 5(3x+4y) = 15x+20y = 5M; (-3)(5x+6y) = -15x-18y = -3N. Add the two equations: 2y = 5M-3N —> y = (1/2)(5M-3N).

Now, substitute this into the first equation to obtain x: 3x + 4(1/2)(5M-3N) = 3x +10M-6N = M —> 3x = 6N - 9M —> x = 2N - 3M.

Our ordered pair is thus (x = 2N - 3M, y = (5/2)M -(3/2)N).
—————————————————————–
3.) a.) Let R be the radius of the circle with area A. Recall that A = pi*R^2. Equivalently, R = sqrt(A / pi). Hence, the diameter D = 2*sqrt(A/pi).

We know that the perimeter P = 4S, where S is the side length of the square with area B. Then, D = P = 2*sqrt(A/pi) = 4S. So, S = (1/2)*sqrt(A/pi).

Now, B = S^2. Then, B = (1/4)*(A/pi) = A/(4*pi). Hence, A/B = 4*pi.

b.) From a.), we know that B/A = 1/(4*pi). The sum S is a geometric series since 0 < 1/(4*pi) < 1. The resulting sum is S = 1/(1-B/A) = 1/(1-(1/4*pi)) = (4*pi) / (4*pi -1).
—————————————————————–
4.) a.) For an arbitrary quadratic polynomial r(x) = a(x-p)(x-q) = ax^2 -a(p+q)x +apq = Ax^2+Bx+C, we can see that the sum of the zeroes (p+q) = -B/A.

For our polynomials, we can take the independent sum of roots for both p(x) and q(x) and add them to obtain the sum of roots for p(x)q(x).

For p(x), we have the sum -b/a, and for q(x), we have the sum -e/d. Hence, the sum of roots for p(x)q(x) is -(b/a + e/d)

b.) Now, if x = 3 is a zero for both p(x) and q(x), we can easily obtain the other zeroes. Each zero of p(x) is a zero of p(x)q(x), likewise for q(x).

For p(x), Y1 + Y2 = -b/a = 3 + Y2 —> Y2 = -3-b/a.

For q(x), X1 + X2 = -e/d = 3 + X2 —> X2 = -3-e/d.

Therefore, the four roots of p(x)q(x) are 3, 3, -3-b/a, and -3-e/d.
——————————————————————-
5.) a.) To find the average value of a function f on an interval [a,b] we integrate f over the interval and divide by the length of the interval b-a.

First, we integrate: int [a, b] x ln (x^2) dx. Make the substitution u = x^2 ((1/2)du = x dx). Then, we have (1/2) int [a,b] ln(u) du.

Using integration by parts, one easily obtains (1/2) (u ln(u) -u ) |[a,b] = (1/2) (a ln(a) -a -b ln(b) + b) = (1/2) (ln((a^a)/(b^b)) + (b-a)).

Dividing by the length of the interval, we finally obtain the average value (1/2) ((1/(b-a))ln((a^a)/(b^b)) + 1).

b.) Standard limiting algorithms do not apply here since we obtain an indeterminate form 0*(-infinity) by naively substituting in x=0.

Apply L'Hospital's Rule to ln(x^2) / (1/x) = 2ln(x) / (1/x) to obtain lim(x–>0) (2/x) / (-1/x^2) = lim(x–>0) (-x^2/x) = 0. f(x) tends to zero as x approaches 0.

c.) This is an elementary application of differentiation. Calculate the derivative of f(x) with respect to x: f'(x) = 2ln(x) +2.

At x = 5, the slope is thus f'(5) = 2ln(5)+2. The value of the function f(x) at x=5 is f(5) = 10*ln(5).

Recall the slope-intercept form of a line: y=mx+b —> 10*ln(5) = 2(ln(5)+1)(5) +b —> 10*ln(5) -10*ln(5)-10 = b = -10.

The equation of the tangent line is therefore y = 2(ln(5)+1)x -10.

Posted on 15 September 2016 - 06:47 PM

I don't understand number one at all

Posted on 15 September 2016 - 06:49 PM

Posted on 15 September 2016 - 07:00 PM

I thought I graduated….

Posted on 15 September 2016 - 08:26 PM

Calc in 8th grade… l0l no

Posted on 15 September 2016 - 08:50 PM

OK, OK, I lied. Grade 9.

Posted on 15 September 2016 - 09:06 PM

If you guys are going to post random things like 1+1=2, I'm going to delete your post. This guy is asking for help, stop using this as an excuse to shit post.