Pls answer asap .

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pka said:
\((x-h)^2+(y-k)^2=R^2\) is a circle that is centred at \((h,k)\) that has radius \(R>0\).
Thus with this problem we have
\(x^2+y^2+16y+k=0\\(x-0)^2+(y+8)^2-64+k=0\\x^2+(y+8)^2=64-k\)
Now we are told that the radius is \(10\) , so what?
Click to expand...
I need to find answer for c part??????????
 
Beer soaked comment/query follows.
Nightshade7456 said:
pka said:
\((x-h)^2+(y-k)^2=R^2\) is a circle that is centred at \((h,k)\) that has radius \(R>0\).
Thus with this problem we have
\(x^2+y^2+16y+k=0\\(x-0)^2+(y+8)^2-64+k=0\\x^2+(y+8)^2=64-k\)
Now we are told that the radius is \(10\) , so what?
Click to expand...
I need to find answer for c part??????????
Click to expand...
The table's been set up for you.
Food is on the table.
What're you waiting for?
 
Beer soaked hint follows.
Nightshade7456 said:
Here the wrkings
Click to expand...
Incorrect as noted in post #12.
Requirement (a) was already worked out for you by pka at post #5.
For requirement (b), you are asked to solve for k.
The problem states that the radius is 10.
You should first determine what the value of k is by equating 64 - k with [MATH]10^2[/MATH], or solve for k in [MATH]64 - k = 10^2[/MATH]After you do that, more hints will come your way.
 
tn
jonah2.0 said:
Beer soaked hint follows.

Incorrect as noted in post #12.
Requirement (a) was already worked out for you by pka at post #5.
For requirement (b), you are asked to solve for k.
The problem states that the radius is 10.
You should first determine what the value of k is by equating 64 - k with [MATH]10^2[/MATH], or solve for k in [MATH]64 - k = 10^2[/MATH]After you do that, more hints will come your way.
Click to expand...
tnx got it
 
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