Start with the equation of a circle whose center is at (h,k) and whose radius is r:
(x-h)^2 + (y-k)^2 = r^2
Substituting the given coordinates of the center:
(x-5)^2 + (y-[-5])^2 = r^2, or (x-5)^2 + (y+5)^2 = r^2
Substituding the given coordinates of a point (6,-2) on the circle:
(6-5)^2 + (-2+5)^2 = r^2
Simplifying:
1^2 + 3^2 = r^2, or 1 + 9 = r^2, or 10= r^2. Then r = sqrt(10).
Answer:
4th term is 274.4 cm
5th term is 192.08 cm
Step-by-step explanation:
Here, we want to get the 4th and 5th term
Let us get the common ratio;
That would be ;
392/560 = 560/800 = 0.7
The 4th term will be ar^3
= 800 * 0.7^3 = 274.4
The 5th term will be ar^4
= 800 * 0.7^4 = 192.08
Let 
. The gradient of 
at the point (1, 0, 0) is the normal vector to the surface, which is also orthogonal to the tangent plane at this point.
So the tangent plane has equation
Compute the gradient:
Evaluate the gradient at the given point:
Then the equation of the tangent plane is
Answer:
Step-by-step explanation:
9^-5 = 9^4 x 9^-9
4-9 = -5
6 divided by 4(3/5)
After multiplying the fraction we get 23/5
Dividing a fraction will result to multiplying
6 * 23/5
138/5 is the answer.
