Let's present the given equation first. Deciphering the given code, I think the equation is (n+1)²/n+23. Then, we want to find the maximum value of n. Suppose the complete equation is:
f(n) = (n+1)²/n+23
To find the maximum,let's apply the concepts in calculus. The maxima can be determined by setting the first derivative to zero. Therefore, we use the chain rule to differentiate the fraction. For a fraction u/v, the derivative is equal to (vdu-udv)/v².
f'(n) = [(n+23)(2)(n+1)-(n+1)²(1)]/(n+23)² = 0
[(n+23)(2n+2) - (n+1)²]/(n+23)² = 0
(2n²+2n+46n+46-n²-2n-1)/(n+23)²=0
n²+46n+45=0
n = -1, -45
There are two roots for the quadratic equation. Comparing the two, the larger one is -1. Thus, the maximum value of n is -1.
Answer:
(x-8)^2+(y-16)^2 = 4
Step-by-step explanation:
The equation of a circle is given by
(x-h)^2+(y-k)^2 = r^2 where (h,k) is the center and r is the radius
(x-8)^2+(y-16)^2 = 2^2
(x-8)^2+(y-16)^2 = 4
I would say B because it only makes since, colleges are looking for whats better in students and you will also work in groups also. So i would go with B :)
<h3>
Answer: k = 0 or k = 3</h3>
Explanation:
If you have repeated x values, then you wont have a function. For example, the points (1,5) and (1,6) mean we don't have a function since the input x = 1 leads to multiple outputs y = 5 and y = 6 simultaneously. For a function to be possible, we must have any input lead to exactly one output only.
What we do is set the x coordinates (k^2 and 3k) equal to each other and solve for k
k^2 = 3k
k^2-3k = 0
k(k-3) = 0 .... factoring
k = 0 or k-3 = 0 .... zero product property
k = 0 or k = 3
If k = 0, then (k^2,5) becomes (0,5). Also, (3k,6) becomes (0,6). The two points (0,5) and (0,6) mean the graph fails the vertical line test.
Similarly, if k = 3, then (k^2,5) becomes (9,5) and (3k,6) = (9,6). Another vertical line test failure happens here to show we don't have a function.
If I saw that correctly each position that's your multiplier so look at your 3rd position and make the 3 into a 30 so it would be 270 :)
