will you still come today? Please, is this correct?

yes, it's correct

Expert answered|selymi|Points 2997|

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Asked 5/25/2011 3:18:32 AM

Updated 5/25/2011 4:10:22 AM

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Izuchukwu Okechukwu

thanks weegy. I want you to solve this mathematics problem for me.

Added 5/25/2011 4:10:22 AM

please when will june/july neco exams commenses?

Weegy: To get your neco ssce 2011 exam timetable visit this link [ [ ] ] (More)

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Asked 5/22/2011 2:46:17 PM

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when will june/july neco will start?

Weegy: They did not release any dates yet on when will it start User: who will challange john cena tonight? Weegy: not me at least, sorry mate. I don't know the answer. you can search it in WWF official website, may be (More)

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Asked 5/22/2011 4:42:16 PM

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who will fase randy orton at the over the limit match?

Weegy: mark henry (More)

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Asked 5/22/2011 5:26:31 PM

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I am waiting for reply to my question.

Weegy: Hi, what is your question? User: my question is who will face John cena at the over the limit match tonight? Weegy: sorry, I searched for it. I found no info about it (More)

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Asked 5/22/2011 5:57:06 PM

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if 9 power 1-x is equll to 27 power y. And x-y is equll to minus one whole number one over two. Find the value of x plus y.

Weegy: If 9^(1-x) = 27^y (More)

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Updated 5/22/2011 10:17:38 PM

1 Answer/Comment

I am so sorry for this. Unfortunately, Experts can't hit the enter button or it sends the answer automatically and even more unfortunate is that the conversation closed before I could help you.

Let's do this problem now.

9^(1-x) = 27^y

x - y = -1 1/2... however, I'm going to write this as -3/2 because mixed fractions are just messy.

x - y = -3/2

So let's solve this son of a fun math problem...

We first need to take the log of both sides to bring the exponents down:

log( 9^(1-x) ) = log( 27^y)

(1 - x)*log(9) = y*log(27)

Now get y alone by dividing by log(27):

(1 - x)*log(9)/log(27) = y

log(9)/log(27) = 2/3

(1 - x)(2/3) = y

-(2/3)x + (2/3) = y

Now let's solve for x:

x - y = -3/2

y = x + 3/2

-(2/3)x + (2/3) = x + (3/2)

-(5/3)x = 5/6

x = -1/2

This means we can find y:

y = -1/2 + 3/2

y = 1

So if x = -1/2 and y = 1... x + y = -1/2 + 1 = 1/2

x + y = 1/2

And just to check if x and y are correct, plug back into the original:

9^(1 - x) = 27^y

9^(1-(-1/2)) = 27^1

9^(1 + 1/2) = 27

9^(3/2) = 27

3^(3) = 27

27 = 27

Phew, and for a minute there I thought we were going to have TOO much fun (:

Let's do this problem now.

9^(1-x) = 27^y

x - y = -1 1/2... however, I'm going to write this as -3/2 because mixed fractions are just messy.

x - y = -3/2

So let's solve this son of a fun math problem...

We first need to take the log of both sides to bring the exponents down:

log( 9^(1-x) ) = log( 27^y)

(1 - x)*log(9) = y*log(27)

Now get y alone by dividing by log(27):

(1 - x)*log(9)/log(27) = y

log(9)/log(27) = 2/3

(1 - x)(2/3) = y

-(2/3)x + (2/3) = y

Now let's solve for x:

x - y = -3/2

y = x + 3/2

-(2/3)x + (2/3) = x + (3/2)

-(5/3)x = 5/6

x = -1/2

This means we can find y:

y = -1/2 + 3/2

y = 1

So if x = -1/2 and y = 1... x + y = -1/2 + 1 = 1/2

x + y = 1/2

And just to check if x and y are correct, plug back into the original:

9^(1 - x) = 27^y

9^(1-(-1/2)) = 27^1

9^(1 + 1/2) = 27

9^(3/2) = 27

3^(3) = 27

27 = 27

Phew, and for a minute there I thought we were going to have TOO much fun (:

Added 5/22/2011 10:17:38 PM

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