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Contact [7]
2 years ago
9

Find the difference.

Mathematics
1 answer:
densk [106]2 years ago
7 0

Answer:

B. 13m+8

Step-by-step explanation:

(20m+3)-(7m-5)

= 20m+3-7m+5

= 13m+8

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Nikitich [7]
/frac{0.8}{30}


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2 \div 75
5 0
3 years ago
Si un proyectil asciende verticalmente, y después de 3 segundos alcanza su altura máxima, calcule la velocidad que lleva a la mi
lys-0071 [83]

Answer:

The speed is 20.8  m/s

Step-by-step explanation:

If a projectile ascends vertically, and after 3 seconds it reaches its maximum height, calculate the velocity that it carries to the middle of its downward trajectory

Let the maximum height is h and initial velocity is u.

From first equation of motion

v = u + at

0 = u - g x 3

u = 3 g.....(1)

Use third equation of motion

v^2 = u^2 - 2 gh \\\\0 = 9 g^2 - 2 gh \\\\h = 4.5 g

Let the speed at half the height is v'.

v^2 = u^2 + 2 gh \\\\v'^2 = 0 + 2 g\times 2.25 g\\\\v'^2 = 4.5\times 9.8\times9.8\\\\v' = 20.8 m/s

8 0
2 years ago
Please helpppppppppppppppppppppppppppppppppppppp
kherson [118]
The answer to this one would be B
7 0
3 years ago
Space Shuttle has three computers. Computer A is a primary computer and Computer B and Computer C are auxiliary computers. There
sweet-ann [11.9K]

Answer:

Required probability equals 0.18%

Step-by-step explanation:

The probability that the primary and one auxiliary computer fails equals

1) Probability that A and B fails

2)Probability that A and C fails

Thus required probability equals

P(E)=P(1)(2)+P(1)P(3)\\\\P(E)=0.03\times 0.03+0.03\times 0.03P(E)=0.18%

3 0
3 years ago
A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line
Nataliya [291]

Answer:

Volume is 2000\pi\ m^{3}

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d              (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d                  (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

c = \frac{1}{5}

Therefore,

h(y) = \frac{1}{5}y + 5

Now, the Volume of the pool is given by:

V = \int h(y)dA

where

A = r\theta

A = rdr\ d\theta

Thus

V = \int (\frac{1}{5}y + 5)dA

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta

V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta

V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}

V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}

7 0
3 years ago
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