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Ksju [112]
4 years ago
11

The function g(x) represents f(x)=9 cos (x-pi/2)+3 after translating pi/6 units left and 4 units up. which equation represents g

(x)?
g(x)= 9 cos (x - 2pi/ 3) -1

g(x)= 9 cos (x - 2pi/3) +7

g(x)= 9 cos (x - pi/3) +7

g(x)= 9 cos (x - pi/3) -1
Mathematics
1 answer:
shtirl [24]4 years ago
6 0

Answer:

g(x)= 9 cos (x - \frac{\pi}{3}) +7

Step-by-step explanation:

What is given is f(x) = g cos (x - pi/2) + 3

Note that the standard form of cosine function is a cos (bx + c) + d

a= amplitude

-c/b = phase shift

d = vertical shift

After moving pi/6 to the left --

x = pi/2 - pi/6 = pi/3

once moving pi/6 left, it is at pi/3

moving 4 units up just means adding 4 + 3, and that equals 7, that would be the vertical shift once applied.

So the equation that represents g(x) is C. 9 cos (x - pi/3) + 7

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In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates
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Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

S = ut + \frac{1}{2} at^{2}

Where S is distance

u is the initial velocity

a is acceleration

and t is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

S = S₁

u = 0 m/s ( The player starts at rest)

a = 2.2 m/s²

Then,

S₁ = 0(t) + \frac{1}{2} (2.2)t^{2}

S₁ = 1.1t^{2}

t^{2} = \frac{S_{1} }{1.1}

For player 2

S = S₂

u = 0 m/s ( The player starts at rest)

a = 1.7 m/s²

Then,

S₂ = 0(t) + \frac{1}{2} (1.7)t^{2}

S₂ = 0.85t^{2}

t^{2} = \frac{S_{2} }{0.85}

Since the time spent by both players is equal, We can  write that

t^{2} = t^{2}

Then,

\frac{S_{1} }{1.1} =  \frac{S_{2} }{0.85}

0.85S_{1} = 1.1S_{2}

From, S₁ + S₂ = 41 m

S₂ = 41  - S₁

Then,

0.85S_{1} = 1.1 ( 41 -S_{1} )

0.85S_{1} = 45.1 - 1.1S_{1}

0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1}  = 23.13 m

This is the distance covered by the first player. We can then put this value into t^{2} = \frac{S_{1} }{1.1} to determine how much time elapses before the players collide.

t^{2} = \frac{S_{1} }{1.1}

t^{2} = \frac{23.13 }{1.1}

t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs

Hence, the time that elapses before the players collide is 4.59 secs

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