basic algebra and arithmetic "marathon". (2 Viewers)

ohexploitable · Apr 7, 2010

New Question:

Solve:

$5^{2x-1}-6(5^{x-1})+1=0$

hscishard · Apr 7, 2010

ohexploitable said:
I wish I paid attention in class

Did you get:

$\begin{align*}\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta&=\frac{4}{9}\\\left(-\frac{b}{a}\right)^{2}-2\left(\frac{c}{a}\right)&=\frac{4}{9}\end{align*}$

???

Yeah.
OMG YOUR DOING THAT ALREADY?

Thank god I did it before everyone else

hscishard · Apr 7, 2010

ohexploitable said:
New Question:

Solve:

$5^{2x-1}-6(5^{x-1})+1=0$

why did it have to be 5^2x-1 and not 5^2x-2

hscishard · Apr 7, 2010

Random x = 1. Guess and check Lol

ohexploitable · Apr 7, 2010

hscishard · Apr 7, 2010

ohexploitable said:
$5^{2x-1}-6(5^{x-1})+1&=0\\\frac{5^{x^{2}}}{5}-6\left(\frac{5^{x}}{5}\right)+1=0\\5^{x^{2}}-6\cdot5^{x}+5=0\\\\\text{let }5^{x}=v\\\\v^{2}-6v+5=0\\(v-1)(v-5)=0\\v=1\text{ or }5\\\\\therefore5^{x}=1\text{ or }5\\\therefore x=0\text{ or }1$

Nice

ohexploitable · Apr 7, 2010

hscishard said:
Yeah.

$\begin{align*}\left(-\frac{b}{a}\right)^{2}-2\left(\frac{c}{a}\right)&=\frac{4}{9}\\\left(-\frac{2}{3}\right)^{2}-2\left(\frac{2-p}{3}\right)&=\frac{4}{9}\\-2\left(\frac{2-p}{3}\right)&=0\\p&=2\end{align*}$

What am I doing wrong???

hscishard · Apr 7, 2010

b is 3.

Unless you wrote the question wrong on the forum. Lol

Amogh · Apr 7, 2010

ohexploitable said:
$5^{2x-1}-6(5^{x-1})+1&=0\\\frac{5^{x^{2}}}{5}-6\left(\frac{5^{x}}{5}\right)+1=0\\5^{x^{2}}-6\cdot5^{x}+5=0\\\\\text{let }5^{x}=v\\\\v^{2}-6v+5=0\\(v-1)(v-5)=0\\v=1\text{ or }5\\\\\therefore5^{x}=1\text{ or }5\\\therefore x=0\text{ or }1$

$5^{2x-1}-6(5^{x-1})+1&=0\\\frac{5^{2x}}{5}-6\left(\frac{5^{x}}{5}\right)+1=0\\{5^{2x}}-6\cdot5^{x}+5=0$

ohexploitable · Apr 7, 2010

hscishard said:
b is 3.

Unless you wrote the question wrong on the forum. Lol

lol, I typed up the question wrong! b is meant to be 2

ohexploitable · Apr 7, 2010

Amogh said:
$5^{2x-1}-6(5^{x-1})+1&=0\\\frac{5^{2x}}{5}-6\left(\frac{5^{x}}{5}\right)+1=0\\{5^{2x}}-6\cdot5^{x}+5=0$

Yar, the 5^x is meant to be in brackets in my post.

ohexploitable · Apr 7, 2010

New Question:

A garment manufacturer calculated the cost $C of making d dresses per day to be given by the formula C = x² - 24x + 250

(i) How many dresses should be manufactured in a day to minimize the production cost?

(ii) Calculate the minimum daily cost of production.

hscishard · Apr 7, 2010

ohexploitable said:
lol, i typed up the question wrong! B is meant to be 2

:d

Amogh · Apr 7, 2010

(i) 12
(ii) 12² - 24*12 + 250

What a retarded garment manufacturer
Anyway

$Solve & & \frac{5}{x-2} > 1$

hscishard · Apr 7, 2010

ohexploitable said:
New Question:

A garment manufacturer calculated the cost $C of making d dresses per day to be given by the formula C = x² - 24x + 250

(i) How many dresses should be manufactured in a day to minimize the production cost?

(ii) Calculate the minimum daily cost of production.

Lol. What the hell is x?

hscishard · Apr 7, 2010

Amogh said:
(i) 12
(ii) 12² - 24*12 + 250

What a retarded garment manufacturer
Anyway

$Solve & & \frac{5}{x-2} > 1$

Its a 2unit area. Lol

(x-2)(x-7) < 0. x is not 2.

2 <x <7

x is greater than 2 but lesser than 7.

Omg for some reason only 2 comes up. ><.
OMG even latex is going against me

Amogh · Apr 7, 2010

I think he did a typo
x is meant to be d (i.e number of dresses)

thongetsu · Apr 7, 2010

5(x-2)>(x-2)^2
5x - 10>x^2 - 4x + 4
0> x^2 -9x +14
(x-2)(x-7) < 0
x = 2,7
2<x<7

thongetsu · Apr 7, 2010

2 < x < 7

ohexploitable · Apr 7, 2010

hscishard said:
Lol. What the hell is x?

x=-b/2a

basic algebra and arithmetic "marathon". (2 Viewers)

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