basic algebra and arithmetic "marathon". (1 Viewer)

nonsenseTM · Mar 10, 2010

ohexploitable said:
You're meant to post a new question.

New Q.

Simplify:

(x – a)(x – b)(x – c) . . . (x – z)

that should be as least a 5u question

ohexploitable · Mar 10, 2010

nonsenseTM said:
that should be as least a 5u question

As above, the answer is 0 because when you get to (x - x)...

Lolsmith · Mar 10, 2010

No questions posted...? If so:

2x + 5y = 26
7x + 19y = 97

luke42 · Mar 11, 2010

2x + 5y = 26
7x + 19y = 97

14x + 35y = 182 -
(14x + 38y = 194)

14x + 35y - 182 - 14x - 38y + 194 = 0

-3y + 12 = 0
-3y = -12
y = 4

14x + 35(4) = 182
14x +140 = 182
14x = 42
x = 3

14x + 38(4) = 194
14x + 152 = 194
14x = 42
x = 3

Q.

find the intergral of 10^x

mirakon · Mar 11, 2010

This is basic algebra and arithmetic, NOT integration.

ohexploitable · Mar 11, 2010

luke42 said:
2x + 5y = 26
7x + 19y = 97

14x + 35y = 182 -
(14x + 38y = 194)

14x + 35y - 182 - 14x - 38y + 194 = 0

-3y + 12 = 0
-3y = -12
y = 4

14x + 35(4) = 182
14x +140 = 182
14x = 42
x = 3

14x + 38(4) = 194
14x + 152 = 194
14x = 42
x = 3

Q.

find the intergral of 10^x

This is for 2011ers

luke42 · Mar 11, 2010

oh my bad.

Q.

Find the first derivative of (x+1)^2 (x+2)^2

Drongoski · Mar 11, 2010

mirakon said:
(2^(2/3)+ 2^(1/3) +1)

Shouldn't the middle term be a + ?

Lolsmith · Mar 11, 2010

I don't think 2011'ers have done Calculus either.... not basic algebra or arithmetic.

mirakon · Mar 12, 2010

Drongoski said:
Shouldn't the middle term be a + ?

Yeah, my bad.

ohexploitable · Mar 14, 2010

$-1<x\leq3$

Find the range of values of:

$x^{2}+3$

hscishard · Mar 14, 2010

ohexploitable said:
$-1<X\LEQ3[ p tex]<> Find the range of values of: [tex]x^{2}+3$

Is it greater than or equal to 3?

I think I should post a question for incase.
Solve for x
(x+2)/(x+1)^2= 1

ohexploitable · Mar 14, 2010

hscishard said:
Is it greater than or equal to 3?

I don't know either

Bob says it's:

$3\leq x^{2}+3\leq12$

I think I should post a question for incase.
Solve for x
(x+2)/(x+1)^2= 1

$\frac{-1\pm\sqrt{5}}{2}$

Something tells me it's not this simple...

ohexploitable · Mar 21, 2010

New Question:

$\sqrt{15-6\sqrt{6}}=x-\sqrt{y}$

Find x and y.

bouncing · Mar 23, 2010

ohexploitable said:
New Question:

$\sqrt{15-6\sqrt{6}}=x-\sqrt{y}$

Find x and y.

how :|

ohexploitable · Mar 23, 2010

bouncing said:
how :|

says a 2010er... lol

$\begin{align*}\sqrt{15-6\sqrt{6}}&=x-\sqrt{y} \\15-6\sqrt{6}&=x^{2}-2x\sqrt{y}+y \\ \\2x&=6 \\y&=6 \\ \\\therefore x=3,\;y=6 \end{align*}$

bouncing · Mar 23, 2010

lol what

i still dont get it

ohexploitable · Mar 23, 2010

lol

http://community.boredofstudies.org...ion-1/139151/equality-surdic-expressions.html

hscishard · Mar 24, 2010

May help if you rearrange it
$15-6\sqrt{6}= (x^{2}+y) - 2x\sqrt{y}$
Then
$15 = x^{2} + y$
and
$6 = 2x$
$\sqrt{y} = \sqrt{6}$
y = 6
x = 3.

Which is true if you sub it back in.

bouncing · Mar 24, 2010

hscishard said:
May help if you rearrange it
$15-6\sqrt{6}= (x^{2}+y) - 2x\sqrt{y}$
Then
$15 = x^{2} + y$
and
$6 = 2x$
$\sqrt{y} = \sqrt{6}$
y = 6
x = 3.

Which is true if you sub it back in.

ahh okay!
that makes alot of sense thanks

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