Well you have runner 1 in lane 1. runner 1 in lane 2 and so on. so it becomes 1,1 1,2 1,3 1,4 1,5 1,6 and then 2,1 2,2 2,3 2,4 2,5 2,6 then 3, and so on
To find the amplitude and period you need to be familiar with the following equation. Also you need to know that the standard cos has a period of

and the midline is a line that runs between the max and min of the y-values of the function.
Equations:f(x) = A cos(Bx +C) + D
f(x) = -4 cos(2x -n) + 3
A = amplitude = |-4| = 4
B = 2
C = phase shift = n = 0
D = vertical shift = midline = 3
Amplitude = 4
Find the period:
Find the midline:We know that the amplitude is 4 so we have a range from -4 to 4. The standard y = cos(x) has its midline at 0 so y = 0. This is also true for y = -4 cos(x). In your equation though, you have a vertical shift of +3 so this changes our midline. With an amplitude of 4, which gives us a range from -4 to 4(our y-values), the shift moves this up by 3 so that means we will have new
y-values and a range of -1 to 7. Now we need to find the midline(
the middle of our y-values) of our new range. We can find this by using the following formula
Midline:y = 3
Note, in the following equations that D = 3 = midliney = A (Bx+C) + D
y = -4 (2x + n) + 3
Also, the picture that is attached is what your equation looks like when graphed.
Answer: 8/3
Step-by-step explanation:
Answer:
x = 4
Therefore, for the triangles to be congruent by HL, the value of x must be 4.
Step-by-step explanation:
Given: ΔABC and ΔHGL are congruent. ∠ABC = ∠HGL = 90°.
Length of hypotenuse AC = 15
Length of hypotenuse HL = 3x + 3
Length of AB = 9, Length of BC = 12 and Length of GL = 2x + 1.
Sol: ∵ ΔABC ≅ ΔHGL
Length of HL = Length of AC (corresponding parts of congruent triangles)
3x + 3 = 15
3x = 15 - 3
3x = 12
x = 12/3 = 4
Therefore, for the triangles to be congruent by HL, the value of x must be 4.
First add liile terms so you equation will be 4x+7=16, then subtract 7 from both sides and you get 4x=9, finally divide both sides by 4 and you will get x=2.25