Note that <span>x^2+y^2=16x−26y−133 represents a circle.
x^2 - 16x + y^2 + 26y = 133
x^2 - 16x + 64 - 64 + y^2 + 26y + 169 - 169 = 133 (completing the square)
Then (x-8)^2 - 64 + (y+13)^2 - 169 = 133
Simplifying, (x-8)^2 + (y+13)^2 = 133 + 169 + 64 = 366
This circle acts as a constraint on the value of 6x-8y. Assume that x and y are both on the circle. Just supposing that x = 10, find y:
(10-8)^2 + (y+13)^2 = 366, or 4 + (y+13)^2 = 366, or (y+13)^2 = 362
This is a quadratic equation that could be solved for y, and the result(s) could be subst. into the expression 6x-8y.
If you were to repeat this exercise several times, for different values of x, you'd come up with various values of 6x-8y and in that way approach (if not find) a definite answer to "</span><span>Given that x2+y2=16x−26y−133, what is the biggest value that 6x−8y can have?"
Hope someone else can come up with a more elegant approach.</span>
Answer:
c 
Step-by-step explanation:
![113,391 = 4\frac{9}{10}[2\frac{1}{10}]^{2} + 135 \\ \\ 113 ≈ h](https://tex.z-dn.net/?f=113%2C391%20%3D%204%5Cfrac%7B9%7D%7B10%7D%5B2%5Cfrac%7B1%7D%7B10%7D%5D%5E%7B2%7D%20%2B%20135%20%5C%5C%20%5C%5C%20113%20%E2%89%88%20h)
I am joyous to assist you anytime.
<u>Answer:</u>
The sprinkler will cover 104.66 square feet
<u>Explanation:</u>
A water sprinkler covers a circular area with a diameter D= 20 feet
and radius r = 10 feet
Arc measuring degrees = 120 degree
Area of the Arc = 
= 
= 
= 104.66
The sprinkler will cover an arc area covering 104.66 square feet
A statically defined term. Pi, for instance, is a constant, as it has a single definition. The speed of light is also a constant.
