We have a "rectangular" double loop, meaning that both loops go to completion.
So there are 3*4=12 executions of t:=t+ij.
Assuming two operatiions per execution of the innermost loop, (i.e. ignoring the implied additions in increment of subscripts), we have 12*2=24 operations in all.
Here the number of operations (+ or *) is exactly known (=24).
Big-O estimates are used for cases with a varying scale of operations, governed by a variable (usually n) to indicate the sensitivity of the number of operations relative to a change in the size of n.
Here we do not have a scale, nor n is defined. The number of operations is constant and known at 24. So a variable is required to find the big-O estimate.
The Pythagorean theorem tells you the square of the hypotenuse is equal to the sum of the squares of the sides.
BC² = AB² + AC²
BC² = 4² +4² = 32
BC = √32 = 4√2
BC = 4*1.41 = 5.64
The length of segment BC is 5.64 m.
Answer:
180
Step-by-step explanation:
15% of 300 is 45.
25% of 300 is 75.
45+75 is 120.
300-120 is 180
Original coordinates of the points:
A (8,15) ; B (12,13) ; C (8,10)
Dilated scale factor of 3.
A ⇒ 3x = 3(8) = 24 ; 3y = 3(15) = 45 ⇒ A' (24,45)
B ⇒ 3x = 3(12) = 36 ; 3y = 3(13) = 39 ⇒ B' (36, 39)
C ⇒ 3x = 3(8) = 24 ; 3y = 3(10) = 30 ⇒ C' (24, 30)
The given image forms a right triangle. So, I'll get the short leg and long leg of the right triangle to solve for the hypotenuse, length of CB.
Short leg: y value of B and C
39 - 30 = 9
Long leg: x value of B and C
36 - 24 = 12
a² + b² = c²
9² + 12² = c²
81 + 144 = c²
225 = c²
√225 = √c²
15 = c
The length of CB is 15 units.
