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Maksim231197 [3]
3 years ago
8

How do I solve this equation

Mathematics
2 answers:
kow [346]3 years ago
6 0
2*3.14*R is the equation solve
R=25
the answer is 157.08
worty [1.4K]3 years ago
3 0
\pi Times diameter 

radius is 25 cm 
diameter would be 50 cm 
the. You multiply 3.14 by 50 to get your answer which is 157cm
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C≥12,E≥6,C+E≤26

Step-by-step explanation:

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6. What is the domain of this function?<br> (0, 2), (1, 3), (2, 4), (3,5), (4, 6)
forsale [732]

\large \underline\mathcal{ » \: QUESTION \:  : }

6. What is the domain of this function?

(0, 2), (1, 3), (2, 4), (3,5), (4,6)

the domain is the set of all the values x of while the range is the set of all the value of y

\tt \:1.) (0, 2), (1, 3), (2, 4), (3,5), (4, 6)

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3 years ago
If pp is inversely proportional to the square of qq, and pp is 24 when qq is 7, determine pp when qq is equal to 2.
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6 0
3 years ago
Substitute 5 for x and evaluate the expression below.<br> (x-2) +11
vekshin1

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3 years ago
Kesha threw her baton up in the air from the marching band platform during practice. The equation h(t) = −16t² + 54t + 40 gives
lapo4ka [179]

Answer:

a) 40 feet

b) 54 ft/min

c) 4 mins

Step-by-step explanation:

Solution:-

- Kesha models the height ( h ) of the baton from the ground level but thrown from a platform of height hi.

- The function h ( t ) is modeled to follow a quadratic - parabolic path mathematically expressed as:

                           h ( t ) = −16t² + 54t + 40

Which gives the height of the baton from ground at time t mins.

- The initial point is of the height of the platform which is at a height of ( hi ) from the ground level.

- So the initial condition is expressed by time = 0 mins, the height of the baton h ( t ) would be:

                         h ( 0 ) = hi = -16*(0)^2 + 54*0 + 40

                         h ( 0 ) = hi = 0 + 0 + 40 = 40 feet

Answer: The height of the platform hi is 40 feet.

- The speed ( v ) during the parabolic path of the baton also varies with time t.

- The function of speed ( v ) with respect to time ( t ) can be determined by taking the derivative of displacement of baton from ground with respect to time t mins.

                        v ( t ) = dh / dt

                        v ( t )= d ( −16t² + 54t + 40 ) / dt

                        v ( t )= -2*(16)*t + 54

                        v ( t )= -32t + 54

- The velocity with which Kesha threw the baton is represented by tim t = 0 mins.

Hence,

                        v ( 0 ) = vi = -32*( 0 ) + 54

                        v ( 0 ) = vi = 54 ft / min

Answer: Kesha threw te baton with an initial speed of vo = 54 ft/min

- The baton reaches is maximum height h_max and comes down when all the kinetic energy is converted to potential energy. The baton starts to come down and cross the platform height hi = 40 feet and hits the ground.

- The height of the ball at ground is zero. Hence,

                     h ( t ) = 0

                     0 = −16t² + 54t + 40

                     0 = -8t^2 + 27t + 20

- Use the quadratic formula to solve the quadratic equation:

                     

                    t = \frac{27+/-\sqrt{27^2 - 4*8*(-20)} }{2*8}\\\\t = \frac{27+/-\sqrt{1369} }{16}\\\\t = \frac{27+/-37 }{16}\\\\t =  \frac{27 + 37}{16} \\\\t = 4

Answer: The time taken for the baton to hit the ground is t = 4 mins

3 0
3 years ago
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