Author Topic: Samuch's facepalm question thread (Read 2283 times) Tweet Share

cipherpol · « **Reply #15 on:** January 15, 2010, 01:49:38 pm »

For the second one;

$x+y=28$

so, $y=28-x$

subbing into [2], $x(28-x)=187$

$28x-x^2=187$

$\implies x^2-28x+187=0$

Solve for x.

samuch · « **Reply #16 on:** January 15, 2010, 01:50:20 pm »

Quote from: cipherpol on January 15, 2010, 01:49:38 pm

For the second one;

$x+y=28$

so, $y=28-x$

subbing into [2], $x(28-x)=187$

$28x-x^2=187$

$\implies x^2-28x+187=0$

Solve for x.

thanks

cipherpol · « **Reply #17 on:** January 15, 2010, 01:52:55 pm »

For the first one;

$x^2 + y^2=178$ [1]

$x+y=13$ [2]

so, $y=13-x$

subbing into [1], $x^2 + (13-x)^2 =178$

$x^2 +x^2 -26x+169=178$

$\implies 2x^2 -26x -9=0$

Solve for x.

the.watchman · « **Reply #18 on:** January 15, 2010, 01:53:31 pm »

For the first one:

From eq. 2: $y=13-x$

Sub this into eq. 1: $x^2 + (13-x)^2=178$

$2x^2-26x-9=0$

$x=\frac{\sqrt{187}+13}{2}$ OR $\frac{13-\sqrt{187}}{2}$

sub these values for x into the second equation to get values for y

EDIT: Soz cipherpol, you were too fast!!!

samuch · « **Reply #19 on:** January 15, 2010, 01:54:54 pm »

thanks guys

i was going the wrong way about it

samuch · « **Reply #20 on:** May 08, 2010, 12:35:34 pm »

hey guys i have a problem

i just dont know how to start with this question!

so if someone could show me how to do the first one that would be awesome

the.watchman · « **Reply #21 on:** May 08, 2010, 12:51:58 pm »

For the first one:

$(f o g) (x)' = f'(g(x)) \cdot g'(x)$

So $(f o g) (1)' = f'(g(1)) \cdot g'(1) = f'(-1) \cdot (-2) = 2 \cdot (-2) = -4$

Use this process for the rest (apply the chain, product and quotient rules)

samuch · « **Reply #22 on:** May 08, 2010, 12:55:39 pm »

Quote from: the.watchman on May 08, 2010, 12:51:58 pm

For the first one:

$(f o g) (x)' = f'(g(x)) \cdot g'(x)$

So $(f o g) (1)' = f'(g(1)) \cdot g'(1) = f'(-1) \cdot (-2) = 2 \cdot (-2) = -4$

Use this process for the rest (apply the chain, product and quotient rules)

oh i get it

thank youuu

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