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laiz [17]
3 years ago
11

Can somebody help me i have to drag the functions on top onto the bottom ones to match their inverse functions.

Mathematics
2 answers:
stellarik [79]3 years ago
8 0

Answer:

1. x/5

2. cubed root of 2x

3.x-10

4.(2x/3)-17

Step-by-step explanation:

marusya05 [52]3 years ago
5 0

Answer:

Step-by-step explanation:

1. Lets find the inverse function for function f(x)=2*x/3-17

To do that first express x through f(x):

2*x/3= f(x)+17

2*x=(f(x)+17)*3

x=(f(x)+17)*3/2   done !!!                        (1)

Next : to get the inverse function from (1) substitute x by f'(x)   and f(x) by x.

So the required function is f'(x)=(x+17)*3/2 or f'(x)=3*(x+17)/2

This is function is No4 in our list. So f(x)=2*x/3-17 should be moved to the box No4  ( on the bottom) of the list.

2.  Lets find the inverse function for function f(x)=x-10

To do that first express x through f(x):

x= f(x)+10

x=f(x)+10   done !!!                        (2)

Next : to get the inverse function from (2) substitute x by f'(x)   and f(x) by x.

So the required function is f'(x)=x+10

This is function is No3 in our list. So f(x)=x-10 should be moved to the box No3  ( from the top) of the list.

3.Lets find the inverse function for function f(x)=sqrt 3 (2x)

To do that first express x through f(x):

2*x= f(x)^3

x=f(x)^3/2   done !!!                        (3)

Next : to get the inverse function from (3) substitute x by f'(x)   and f(x) by x.

So the required function is f'(x)=x^3/2

This is function No2 in our list. So f(x)=sqrt 3 (2x) should be moved to the box No2  ( from the top) of the list.

4.Lets find the inverse function for function f(x)=x/5

To do that first express x through f(x):

x=f(x)*5   done !!!                        (4)

Next : to get the inverse function from (4) substitute x by f'(x)   and f(x) by x.

So the required function is f'(x)=x*5 or f'(x)=5*x

This is function No1 in our list. So f(x)=x/5 should be moved to the box No1  ( on the top) of the list.

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Read 2 more answers
Luiza is jumping on a trampoline.
lyudmila [28]

Answer:

The second time when Luiza reaches a height of 1.2 m = 2 08 s

Step-by-step explanation:

Complete Question

Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts jumping. Here, the angle is entered in radians.

H(t) = -0.6 cos (2pi/2.5)t + 1.5.

What is the second time when Luiza reaches a height of 1.2 m? Round your final answer to the nearest hundredth of a second.

Solution

Luiza is jumping on trampolines and her height above the levelled ground at any time, t, is given as

H(t) = -0.6cos⁡(2π/2.5)t + 1.5

What is t when H = 1.2 m

1.2 = -0.6cos⁡(2π/2.5)t + 1.5

0.6cos⁡(2π/2.5)t = 1.2 - 1.5 = -0.3

Cos (2π/2.5)t = (0.3/0.6) = 0.5

Note that in radians,

Cos (π/3) = 0.5

This is the first time, the second time that cos θ = 0.5 is in the fourth quadrant,

Cos (5π/3) = 0.5

So,

Cos (2π/2.5)t = Cos (5π/3)

(2π/2.5)t = (5π/3)

(2/2.5) × t = (5/3)

t = (5/3) × (2.5/2) = 2.0833333 = 2.08 s to the neareast hundredth of a second.

Hope this Helps!!!

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= (4x + 1) + (x² - 5)

= x² + 4x + (1 - 5)

= x² + 4x - 4

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