vash_08

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vash_08's Profile Comments

Comments: 7, viewing 1 - 7

GNiCk89 wrote on Aug 6th, 2009 6:06am

Yeeeaaah

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radomu wrote on May 28th, 2009 3:02am

Four Dead in Ohio! Gotta Get Down To It, Soldiers Are Cutting Us Down!

sorry

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brandon369852 wrote on Jan 3rd, 2009 5:46pm

vash_08 wrote on Jan 3rd, 2009 at 12:52am :

Hey man, you might not remember, but early December you helped me out with some Algebra, I finally got my exam grade, and I got a B+, which was pretty high compared to my high school algebra grades, so I wanted to thank you for the help!

Always glad to help!

And I'm banned, but if you ever have any other questions, just let me know, and I'll be glad to help again.

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hippotato7 wrote on Jan 3rd, 2009 5:15pm

vash_08 wrote on Jan 3rd, 2009 at 1:42am :

Nice moustache bro. OH

IO

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brandon369852 wrote on Dec 9th, 2008 12:59am

Now for the next one. It reads:

10a^2 + 43a + 28

You were right in multiplying a and c(the constants), so you know how to do that.

So you got: 10a^2 + 8a + 35a + 28

You grouped them like you said:

(10a^2 + 8a) + (35a + 28)

Find like terms in both factors:

2a(5a + 4) + 7(5a + 4)

Note that the two factors in the brackets(5a + 4), so they are able to factor.

Next, you'll just do the same as above.

(2a + 7) (5a + 4)

And that's fully factored because you can't take anything else out from either one.

Hope I helped in time for your exams.

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brandon369852 wrote on Dec 9th, 2008 12:55am

I saw you needed help in the maths thread, but I couldn't reply because I'm banned.

I'll show you how to do the second and the third one here.

I understand that the second question asked:
27b^2 + 6bx - 63b - 14x

You said to solve by grouping, which means you just put brackets around the two pairs.

(27b^2 + 6bx) - (63b - 14x)

Now take out common terms from both:

3b(9b + 2x) - 7(9b + 2x)

Note that the two brackets(9b+2x) are the same; this means they are able to factor.

So you'll take the two factors:

(3b - 7) (9b + 2x)

Those are fully factored, but if your teacher asks for you to solve them, let me know and I'll show you how to do that(basically set both factors to zero and then solve)

If that makes sense, than woo, if not, comment me back and I'll go over it more in depth.

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Arcticallica wrote on Dec 8th, 2008 10:07pm

Hey man i posted a reply to your algebra question --

The answers should be (6z-12)(z-6) and (5a+4)(2a+7).
If you expand the brackets you get what you put so i presume that's what you're after? But I'm no expert mate!

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