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kogti [31]
3 years ago
8

[2.4–(0.3–3.21)÷2+0.44÷(−2)]÷ 2/5

Mathematics
1 answer:
netineya [11]3 years ago
6 0

Answer:9.0875

Step-by-step explanation:

[2.4–(0.3–3.21)÷2+0.44÷(−2)]÷ 2/5=

[4.8:2-(-2.91):2-0.44:2]:2/5=

=[(4.8+2.91-0.44):2]*5/2=

=[(7.71-0.44):2]*5/2=

=(7.27:2)*5/2=(7.27*5):(2*2)=

=36.35:4=9.0875

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Need help got 20min left almost done with class please let me know if you can help me finish this class
yanalaym [24]

Answer:

Linear because every x, it goes -4 in y.  Therefore, it is linear.

Hope this helps :)

5 0
3 years ago
Read 2 more answers
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
Hey!<br> Question: <br> Simplify (16y^8)^3/4<br><br> Thanks for the help!
olasank [31]
Hope this helps! Is from the app Photomath.

8 0
3 years ago
Neeeeed helpppppp please
MissTica

Answer:

Scale factor = 7

Step-by-step explanation:

Dilation with scale factor to map HEFG to DABC will be,

Scale factor = \frac{\text{Dimension of Image DABC}}{\text{Dimension of preimage HEFG}}

                    = \frac{\text{Length of CD}}{\text{Length of GH}}

Length of CD = Distance between two points C(0, -7) and D(-7, 0)

                       = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

                       = \sqrt{(-7-0)^2+(0+7)^2}

                       = \sqrt{98}

                       = 7\sqrt{2}

Length of GH = Distance between G(0, -1) and H(-1, 0)

                       = \sqrt{(-1-0)^2+(0+1)^2}

                       = \sqrt{2}

Scale factor = \frac{7\sqrt{2} }{\sqrt{2} }

                    = 7

6 0
3 years ago
Find the area of the figure.
blondinia [14]
I hope this helpssss :))

5 0
2 years ago
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