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

A: 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

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)

Question

Expert Answered

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

A cyclist starts from a point X and rides 3km due west to a point Y. At Y, he changes direction and rides 5km north-west to a point Z. 1. How far is he from the starting point, correct to the nearest km? 2. Find the bearing of Z from X, to the nearest degree.

Question

Not Answered

Updated 7/22/2013 2:01:06 AM

1 Answer/Comment

(b)(i). Let us take the North-West change in direction at Y as 45 degrees, (this is because, at the change in direction, no angle was given, so take an average of the angle NW i.e. average of 90 degrees)

This implies as Angle Y = 90 + 45 = 135

By cosine rule,

/XZ/^2 = /XY/^2 + /YZ/^2 – 2*/XY/*/YZ/*cos(Y)

Where cos Y = cos 135 degree

Substituting,

/XZ/^2 = 3^2 + 5^2 – 2 × 3 × 5 cos135

But cos1350 = -0.7071

Substituting,

/XZ/^2 = 9 + 25 – (30 × -0.7071)

/XZ/^2 = 34 + 21.213 = 55.213

/XZ/ = 55.213

/XZ/ = 7.4305km

Distance from starting point = /XZ/ = 7km. Answer.

(b)(ii). The bearing,

By sine rule,

(sin )/5 = (sin 135)/7.4305

Sin = (5 × sin135)/7.4305

But sin135 = 0.7071,

Substitute for sin135,

Sin = (5 × 0.7071)/7.4305

Sin = 0.4758

= 0.4758/sin = sin-1(0.4758)

= 28.4115

Bearing of Z from X = N90- W

OR

Bearing of Z from X = 270 +

Substituting,

Bearing of Z from X = 90 – 28.4115 = 61.5885 degree

OR

Bearing of Z from X = 270 + 28.4115 = 298.4115 degree

To the nearest degree

Bearing of Z from X = N62W. Answer.

OR

Bearing of Z from X = 298 degree. Answer

This implies as Angle Y = 90 + 45 = 135

By cosine rule,

/XZ/^2 = /XY/^2 + /YZ/^2 – 2*/XY/*/YZ/*cos(Y)

Where cos Y = cos 135 degree

Substituting,

/XZ/^2 = 3^2 + 5^2 – 2 × 3 × 5 cos135

But cos1350 = -0.7071

Substituting,

/XZ/^2 = 9 + 25 – (30 × -0.7071)

/XZ/^2 = 34 + 21.213 = 55.213

/XZ/ = 55.213

/XZ/ = 7.4305km

Distance from starting point = /XZ/ = 7km. Answer.

(b)(ii). The bearing,

By sine rule,

(sin )/5 = (sin 135)/7.4305

Sin = (5 × sin135)/7.4305

But sin135 = 0.7071,

Substitute for sin135,

Sin = (5 × 0.7071)/7.4305

Sin = 0.4758

= 0.4758/sin = sin-1(0.4758)

= 28.4115

Bearing of Z from X = N90- W

OR

Bearing of Z from X = 270 +

Substituting,

Bearing of Z from X = 90 – 28.4115 = 61.5885 degree

OR

Bearing of Z from X = 270 + 28.4115 = 298.4115 degree

To the nearest degree

Bearing of Z from X = N62W. Answer.

OR

Bearing of Z from X = 298 degree. Answer

Added 7/22/2013 2:01:06 AM

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