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nevsk [136]
3 years ago
6

t;m" t=\frac{ms}{m+n}" alt="solve for "m" t=\frac{ms}{m+n}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

\boxed{\sf \ \ \ m = -\dfrac{tn}{t-s} \ \ \ }

Step-by-step explanation:

Hello,

let s assume that m+n is different from 0

we have this equation and we need to find m as a function of t, s, and n

t=\dfrac{ms}{m+n}

<=>

(m+n)*t=ms\\\\ tm+tn=sm\\ (t-s)m = -tn\\ m = -\dfrac{tn}{t-s}

for t-s different from 0, so t different from s

hope this helps

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What function is shown by the graph below?
Airida [17]

Answer:

The function f(x)=\sqrt{x+2}  is shown by the graph below ⇒ 2nd answer

Step-by-step explanation:

<em>To find the right function chose two points from the graph and substitute the x-coordinate of each point in the function to find the y-coordinate, if they are the same with the corresponding y-coordinates of the points, then the function is shown by the graph</em>

From the figure:

The curve passes through points (-2 , 0) and (2 , 2)

∵  f(x)=\sqrt{x-2}

∵ x = -2

- Substitute x by -2

∴ f(-2)=\sqrt{-2-2}

∴ f(-2)=\sqrt{-4} ⇒ it is impossible no square root for (-) number

∴ f(x)=\sqrt{x-2} is not the function shown by the graph

∵  f(x)=\sqrt{x+2}

∵ x = -2

- Substitute x by -2

∴ f(-2)=\sqrt{-2+2}

∴ f(-2)=\sqrt{0}

∴ f(-2) = 0 ⇒ same as the y-coordinate of x = -2

∵ x = 2

- Substitute x by 2

∴ f(2)=\sqrt{2+2}

∴ f(2)=\sqrt{4}

∴ f(2) = 2 ⇒ same as the y-coordinate of x = 2

∴ The function f(x)=\sqrt{x+2} is shown by the graph below

7 0
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

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3 years ago
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aivan3 [116]
M + 5n = 7
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Sunny_sXe [5.5K]
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