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SuicideFuel Math thread problem (official)

First, x needs to be nonnegative. Let u = the cube root of x. Then sqrt(1 + u^3) = u^2 - 1. Squaring both sides yields that 1 + u^3 = u^4 - 2u^2 + 1 -- i.e., u^4 - u^3 - 2u^2 = 0, so u = 0 or 0 = u^2 - u - 2 = (u - 2)(u + 1). Since u = 0 would imply x = 0, which is evidently a spurious solution, either u = 2 or u = -1. However, u = -1 would mean that x = -1, so u = 2 AKA x = 8 is the only solution.
After DEATH LABOUR CAMP I WILL SOLVE (IN VIDEO GAME)
 
First, x needs to be nonnegative. Let u = the cube root of x. Then sqrt(1 + u^3) = u^2 - 1. Squaring both sides yields that 1 + u^3 = u^4 - 2u^2 + 1 -- i.e., u^4 - u^3 - 2u^2 = 0, so u = 0 or 0 = u^2 - u - 2 = (u - 2)(u + 1). Since u = 0 would imply x = 0, which is evidently a spurious solution, either u = 2 or u = -1. However, u = -1 would mean that x = -1, so u = 2 AKA x = 8 is the only solution.
Correct :feelsokman:
 
@trying to ascend @Ahnfeltia Any book recommendations for a beginner self learning linear algebra?
 
@trying to ascend @Ahnfeltia Any book recommendations for a beginner self learning linear algebra?
The linear algebra book my uni used was in Flemish, so no. The following might be nice, however.
 
Nigga i'm still in high school
Let a be a positive real number. Set S(a) to the value of the enclosed area bounded by the y-axis, by the parabola =x² and by the tangent line to the same parabola at the point (a, a²).


Find the limit: lim→+∞ S(a)/(a³ + a² + a + 1)
 
I have math this semester
 
Absolute value of complex numbers, circumferences in the gauss plane
Ok. A number is 5 times more than twice another number. Find the two numbers and the maximum value.
 
Two numbers have a sum of 29 and a product that is a maximum. Determine the two numbers and the maximum product.
 
PLS HELP ME MY NIGGER HOME WORK WONT GIVE ME THE CORRECT ANSWER
 
Two numbers have a sum of 29 and a product that is a maximum. Determine the two numbers and the maximum product.
Are those integers? If not.

Using the inequality of means, we can see it's 14,5 and 14,5.

If they are integers, then it's 14 and 15
 
Last edited:
Are those integers? If not.

Using the inequality of means, we can see it's 14,5 and 14,5.

If they are integers, then it's 14 and 15
Ye I figured it out but it was partly googles fault because it said X^2 can be negative but it cant
 
Two numbers have a sum of 29 and a product that is a maximum. Determine the two numbers and the maximum product.
While @trying to ascend is right, here's an easy way to see it.

We want to find a & b such that a + b = 90 and a*b is as large as possible. Since a + b = 29, we'll cleverly introduce a new variable c as follows: a = 14.5 + c & b = 14.5 - c. Evidently a + b still equals 29. However, a*b = (14.5 + c)(14.5 - c) = 210.25 - c^2. Since c^2 cannot be negative, the largest value for a*b is achieved when c^2 = 0, i.e., c = 0.
 
What is the method one uses in this kind of problem?
We want to find a & b such that a + b = 90 and a*b is as large as possible. Since a + b = 29, we'll cleverly introduce a new variable c as follows: a = 14.5 + c & b = 14.5 - c. Evidently a + b still equals 29. However, a*b = (14.5 + c)(14.5 - c) = 210.25 - c^2. Since c^2 cannot be negative, the largest value for a*b is achieved when c^2 = 0, i.e., c = 0.
 
Is there a correlation between being good at math and being unattractive to foids
 
Doing maths is cope ai will be here soon
 
Can you guys explain this bullshit to me?

Bullshit
 
Two numbers have a sum of 29 and a product that is a maximum. Determine the two numbers and the maximum product.
Isn't this a Grade 9 question? I thought you were 16:feelswhat:.
 
Isn't this a Grade 9 question? I thought you were 16:feelswhat:.
I'm in Canada so we probably have different educational systems. Also, if its so easy go ahead and solve this:

Two numbers have a difference of
13 and a product that is a minimum.
Determine the two numbers and the
minimum product.
 
I'm in Canada so we probably have different educational systems. Also, if its so easy go ahead and solve this:

Two numbers have a difference of
13 and a product that is a minimum.
Determine the two numbers and the
minimum product.
6,5 and -6,5?
 
yo solve this @Grim_Reaper

What is the maximum total area that
450 cm of string can enclose if it is used
to form the perimeters of two adjoining
rectangles as shown?
 
I'm in Canada so we probably have different educational systems. Also, if its so easy go ahead and solve this:
I live in Canada as well.
Two numbers have a difference of
13 and a product that is a minimum.
Determine the two numbers and the
minimum product.
So let x be the first number and 13+x be the second number
You would have x(13+x)= 13x + x^2
This is a upward parabola so you find the minimum value.
Use the equation -b/2a to find the the x-value of the min value, which is -13/2(2)= -6.5
Then sub the x-value into 13x+x^2, and you get -42.25 as the min product.
 
yo solve this @Grim_Reaper

What is the maximum total area that
450 cm of string can enclose if it is used
to form the perimeters of two adjoining
rectangles as shown?
If it's joining two rectangles, then it would have 3 lengths and 4 widths, so the equation would be 3l + 4w = 450
Area would be l*2w, or 150-4/3w * 2w = 300w - 8/3w^2
x-value of max value is -300/(2(-8/3)) = 56.25
Max value and max perimeter is 8437.5.
 
If it's joining two rectangles, then it would have 3 lengths and 4 widths, so the equation would be 3l + 4w = 450
Area would be l*2w, or 150-4/3w * 2w = 300w - 8/3w^2
x-value of max value is -300/(2(-8/3)) = 56.25
Max value and max perimeter is 8437.5.
@DespressedCurryCel1 if you know calculus, then find the derivative of 300w - 8/3w^2, which is 300-16/3w, then solve for w by setting the equation to 0, and you'll get 56.25, then sub it back into the original equation and you get your answer.
 
@DespressedCurryCel1 if you know calculus, then find the derivative of 300w - 8/3w^2, which is 300-16/3w, then solve for w by setting the equation to 0, and you'll get 56.25, then sub it back into the original equation and you get your answer.
I’m doing pre calculus mang
 

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