Posted on 15 September 2016 - 09:13 PM
LordBowser wrote
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).
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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).
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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).
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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.
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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.
literally no one cares
Posted on 15 September 2016 - 09:13 PM
SporkHandles wrote
thats like 6th grade math…
Posted on 15 September 2016 - 09:13 PM
MeMyselfAndCube wrote
i did pre calc in 8th o.0
Posted on 15 September 2016 - 09:16 PM
ThatOneCombo wrote
MeMyselfAndCube wrote...
i did pre calc in 8th o.0
Posted on 15 September 2016 - 09:23 PM
ThatOneCombo wrote
SporkHandles wrote...
thats like 6th grade math…
tf
Posted on 15 September 2016 - 09:25 PM
ThatOneCombo wrote
LordBowser wrote...
literally no one cares
Haha mad.
Posted on 15 September 2016 - 09:26 PM
Aymbaut wrote
ThatOneCombo wrote...
I mean like in my school Algerbra 2 Honors is avg. in 8th grade and I took some test and skipped pre al so i did al 2 in 7th grade
Posted on 15 September 2016 - 09:29 PM
xIronMatt wrote
Thanks.
Posted on 15 September 2016 - 09:32 PM
ThatOneCombo wrote
MeMyselfAndCube wrote...
i did pre calc in 8th o.0
Posted on 15 September 2016 - 09:33 PM
ThatOneCombo wrote
SporkHandles wrote...
thats like 6th grade math…
Posted on 15 September 2016 - 09:33 PM
PizzqWithAL wrote
ThatOneCombo wrote...
its basic algebra
Posted on 15 September 2016 - 09:33 PM
PizzqWithAL wrote
ThatOneCombo wrote...
??
Posted on 15 September 2016 - 09:34 PM
PizzqWithAL wrote
ThatOneCombo wrote...
What grade is @ThatOneCombo in now?
Posted on 15 September 2016 - 09:34 PM
LordBowser wrote
nine
Posted on 15 September 2016 - 09:34 PM
ThatOneCombo wrote
LordBowser wrote...
nine
What math are you in now? AB or BC? Or neither?
Posted on 15 September 2016 - 09:36 PM
ThatOneCombo wrote
PizzqWithAL wrote...
??
Last edited on 15 September 2016 - 09:36 PM by ThatOneCombo
OH WAIT I'M RETARDED I MEANT I DID ALGEBRA 2 IN 8TH GRADE AND PRE CALC In 9TH KMS @RegalRat @SuperSandwishPosted on 15 September 2016 - 09:37 PM
ThatOneCombo wrote
That's still pretty good and considerably beyond the standard math curriculum at most schools.
Are you planning on doing something math/science-related in college?
Posted on 15 September 2016 - 09:38 PM
ThatOneCombo wrote
Posted on 15 September 2016 - 09:38 PM
LordBowser wrote
ThatOneCombo wrote...
That's still pretty good and considerably beyond the standard math curriculum at most schools.
Are you planning on doing something math/science-related in college?
lol plz no, also yea my school is like 1/2 years ahead. The town right next to me is so shit, the only thing they are good at is track. Not trying to be racist but you can assume what the majority of the school is :P