The total area of pool and border = 361 square feet.
Length of square pool = x
Length of square pool plus border = (x + 1 +1 ) = x + 2
There is 1 foot length on each side on the x length.
The other length of the square pool and border = (x + 2)
Area = (x+2)(x+2)
(x + 2)(x +2) = 361. Let A = x +2
A*A = 361
A² = 361 Take square root of both sides
A = √361
A = 19
A = x + 2 = 19
x = 19 - 2
x = 17
Area of square pool = x*x = 17*17 = 289
Area of the pool = 289 square feet.
Answer:
Since we extract n elements in total, the algorithm for the running time for K sorted list is O (n log k+ k) = O (n log k)
Step-by-step explanation:
To understand better how we arrived at the aforementioned algorithm, we take it step by step
a, Construct a min-heap of the minimum elements from each of "k" lists.
The creation of this min-heap will cost O (k) time.
b) Next we run delete Minimum and move the minimum element to the output array.
Each extraction takes O (log k) time.
c) Then insert into the heap the next element from the list from which the element was extracted.
Now, we note that since we extract n elements in total, the running time is
O (n log k+ k) = O (n log k).
So we can conclude that :
Since we extract n elements in total, the algorithm for the running time for K sorted list is O (n log k+ k) = O (n log k)
You have to find the angle of sunlight and the pole , which then cast a shadow
length of the pole is equal to 3.1m
creating a hypotenuse of 4.52m
the angle will be
[email protected] = opposite side / hypotenuse
= 1.46/4.52
= 0.32 that's 18.66 °
now for the building sin18.66° = opposite/hypotenuse
0.32 = 36.75/ hypotenuse
hypotenuse = 114.84 m
thus using Pythagoras theorem
114.84*114.84 = 36.75*36.75 + x^2
which comes around = 108.8 roughly 109 metres long
Answer:
9·x² - 36·x = 4·y² + 24·y + 36 in standard form is;
(x - 2)²/2² - (y + 3)²/3² = 1
Step-by-step explanation:
The standard form of a hyperbola is given as follows;
(x - h)²/a² - (y - k)²/b² = 1 or (y - k)²/b² - (x - h)²/a² = 1
The given equation is presented as follows;
9·x² - 36·x = 4·y² + 24·y + 36
By completing the square, we get;
(3·x - 6)·(3·x - 6) - 36 = (2·y + 6)·(2·y + 6)
(3·x - 6)² - 36 = (2·y + 6)²
(3·x - 6)² - (2·y + 6)² = 36
(3·x - 6)²/36 - (2·y + 6)²/36 = 36/36 = 1
(3·x - 6)²/6² - (2·y + 6)²/6² = 1
3²·(x - 2)²/6² - 2²·(y + 3)²/6² = 1
(x - 2)²/2² - (y + 3)²/3² = 1
The equation of the hyperbola is (x - 2)²/2² - (y + 3)²/3² = 1.
