Martinez39:
Thanks.

You're welcome!

Martinez39:
How do you and MathsEconomics compose stuff using that white background? Where can I get it?

I use Math Editor while he uses MathMagic Lite. I find both painful to type in.

Martinez39:

Can you find a?
5log₄ a + 48log_a 4 = a/8

a=256

Had to graph it but post your solution if you have one.

Darivie04:

I use Math Editor while he uses MathMagic Lite. I find both painful to type in.

Both are really painful to type in really. I have MathMagic Lite. It's okay but I don't know how to save my work. Still exploring though.

Darivie04:

a=256

Had to graph it but post your solution if you have one.

I tried solving it algebraically but I lost myself.

Hello Mathematicians... Something is making my head to "turnioniown" here... Pls help me out...

Find the Jacobian of transformation from (x, y) to (u, w), defined as:

x + y^2 = u
y + x^2 = w

I am particularly confused as to how to express each of x, y in terms of u, w.

Please help me with this queation
Assuming loge 4.4=1.4816 and love 7.7=2.0142,the value of loge 1/4 is?

Reylwane:
Please help me with this queation
Assuming loge 4.4=1.4816 and love 7.7=2.0142,the value of loge 1/4 is?

Do you mean
log_e 4.4 = 1.4816 and log_e 7.7 = 2.0142

Darivie04:

a=256

Had to graph it but post your solution if you have one.

I'm trying to solve this in algebraic form.

naturalwaves:

I'm trying to solve this in algebraic form.

You're going to need an original idea or a deep observation because straightforward approaches don't seem to go anywhere

Darivie04:

You're going to need an original idea or a deep observation because straightforward approaches don't seem to go anywhere

Yeah. I got stuck someone. Countless substitutions.

Darivie04:

I use Math Editor while he uses MathMagic Lite. I find both painful to type in.

yes,thanks

Martinez39:

Do you mean
log_e 4.4 = 1.4816 and log_e 7.7 = 2.0142

[q

Martinez39:

Do you mean
log_e 4.4 = 1.4816 and log_e 7.7 = 2.0142

uote author=Martinez39 post=81594551]
Do you mean
log_e 4.4 = 1.4816 and log_e 7.7 = 2.0142 [/quote] YES please

Abeg someone should help me with this problem

If (x+1)^3 divides (P(x) - 1) and (x-1)^3 divides (P(x) + 1). Find the fifth degree polynomial P(x).

Darivie04:
Abeg someone should help me with this problem

If (x+1)^3 divides (P(x) - 1) and (x-1)^3 divides (P(x) + 1). Find the fifth degree polynomial P(x).

The fifth degree polynomial is
P(x) = ⅛(-3x⁵ + 10x³ - 15x)

If you are interested in the workings, I could explain.

Martinez39:


The fifth degree polynomial is
P(x) = ⅛(-3x⁵ + 10x³ - 15x)

If you are interested in the workings, I could explain.

I am very interested

Darivie04:

I am very interested

Every polynomial with real coefficient can be factored over the real number into linear factor(s) and/or irreducible quadratic factor(s).

1) From your question,
P(x) - 1 = (x + 1)³(A₁x² + B₁x + C₁) —(1)
P(x) + 1 = (x - 1)³(A₂x² + B₂x + C₁) —(2)

Where A₁, B₂, C₁, A₂, B₂, and C₂ are constants.

Also noticed that
P(-1) = 1 and P(1) = -1

2) Solve eq(1) & eq(2) for P(x) and expand both of them. Then form six equations by matching coefficients. Let's call them the "coefficient" equations. Doing so will lead to

x⁵: A₁ = A₂
x⁴: 3A₁ + B₁ = B₂ - 3A₂
x³: 3A₁ + 3B₁ + C₁ = 3A₂ - 3B₂ + C₂
x²: A₁ + 3B₁ + 3C₁ = 3B₂ - A₂ - 3C₂
x: B₁ + 3C₁ = 3C₂ - B₂
D: C₁ + 1 = -C₂ - 1

3) By cleverly manipulating these equations and substituting in each of them, you will obtain the following solutions
A₁ = A₂ = -(3/8 )
B₁ = - B₂ = -(9/8 )
C₁ = C₂ = -1

You could also have written A₁, A₂, B₁, B₂, C₁, and C₂ as variables and solve the resulting system of linear equations represented by the six "coefficient" equations.

Substituting these solutions in either eq(1) or eq(2) will show that the required polynomial is
P(x) = ⅛(-3x⁵ + 10x³ - 15x)

I’m preparing for GRE exam and I need a very sound private mathematics tutorial teacher in Bodija estate, Ibadan. Kindly drop your contact info or share with any friend who could help with this.. We shall discuss the fee if satisfied with your proficiency in maths upon trial. Please, you must be a wizard in maths/quantitative reasoning.



Thanks

Martinez39:



Every polynomial with real coefficient can be factored over the real number into linear factor(s) and/or irreducible quadratic factor(s).

In your second equation shouldn't that be (x-1)^3

Darivie04:

In your second equation shouldn't that be (x-1)^3

Yes. It is (x - 1)³. Answer is still correct.

Martinez39:



Every polynomial with real coefficient can be factored over the real number into linear factor(s) and/or irreducible quadratic factor(s).

Just had time to go through the solution this morning. At first I was confused because at the same time you dropped your solution someone else dropped a proof that the polynomial doesn't exist. I later found out that the proof just applied to monic polynomials.

I actually had that idea of "six coeffiecient equations" but the size of the system made me quickly turn back.

Anyways, thanks a lot for the help.

Darivie04:

Just had time to go through the solution this morning. At first I was confused because at the same time you dropped your solution someone else dropped a proof that the polynomial doesn't exist. I later found out that the proof just applied to monic polynomials.

I actually had that idea of "six coeffiecient equations" but the size of the system made me quickly turn back.

Anyways, thanks a lot for the help.

Oh well, the solution is not a monic polynomial.

Martinez39:
Oh well, the solution is a monic polynomial.

You mean your solution? The leading coefficient isn't 1.

Darivie04:

You mean your solution? The leading coefficient isn't 1.

I meant to say that the solution exists and it is not a monic polynomial. My solution is correct, why don't you test it?

Though my solution is still correct but I have modified the workings I posted yesterday to make it simpler for you to verify. Check it out.

Darivie04:

You mean your solution? The leading coefficient isn't 1.

The leading coefficient is -⅜.

Martinez39:
I meant to say that the solution exists and it is not a monic polynomial. My solution is correct, why don't you test it?

Though my solution is still correct but I have modified the workings I posted yesterday to make it simpler for you to verify. Check it out.

Your solution is correct. I just thought you we're saying something else.

Greetings, House! It's been a while. Kudos all!

jaryeh:
Hello Mathematicians... Something is making my head to "turnioniown" here... Pls help me out...

Find the Jacobian of transformation from (x, y) to (u, w), defined as:

x + y^2 = u
y + x^2 = w

I am particularly confused as to how to express each of x, y in terms of u, w.

Do you still need help with this?

DanXplore:


Do you still need help with this?

Yes boss, I do.

On Conic Sections.