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4 years ago

Luerityilly CIUSS SECUMS UN SUNIUS

Mathematics

2 answers:

ruslelena [56]4 years ago

5 0

Answer:

cones cylinder sphere

Step-by-step explanation:

GarryVolchara [31]4 years ago

3 0

Answer:

1) cones

3) cylinders

5) spheres

Step-by-step explanation:

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Solving systems

polet [3.4K]

Va rgcdjhf 69:6(;4)$8744;5(333:

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3 years ago

I need hell with this pleasee

olya-2409 [2.1K]

D. last answer
Step by step explanation:

4 0

3 years ago

Plz help got it wrong show work plz

igor_vitrenko [27]

Answer: C) $1951.53

<u>Step-by-step explanation:</u>

$A=P(1+r)^{t}$

A is amount accrued (final balance)
P is the Principal (original amount deposited)
r is the rate (in decimal form)
t is the number of years

A=1500(1+0.068)⁴

= 1500(1.068)⁴

= 1951.53

6 0

4 years ago

A ball is launched upward at 14 m/s from a platform 30 m high.Find the maximum height the ball will reach and how long it will t

BartSMP [9]

Answer:

The ball will reach a maximum height of 39.993 meters after 1.428 seconds.

Step-by-step explanation:

Let suppose that no non-conservative forces acts on the ball during its motion, then we can determine the maximum height reached by the Principle of Energy Conservation, which states that:

$K_{1}+U_{g,1} = K_{2}+U_{g,2}$ (1)

Where:

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies, measured in joules.

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energies, measured in joules.

By definition of translational kinetic energy and gravitational potential energy we expand and simplify the expression above:

$\frac{1}{2}\cdot m\cdot v_{2}^{2}+m\cdot g\cdot y_{2}= \frac{1}{2}\cdot m\cdot v_{1}^{2}+m\cdot g\cdot y_{1}$ (2)

Where:

$m$ - Mass of the ball, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$ , $v_{2}$ - Initial and final speed of the ball, measured in meters per second.

$y_{1}$ , $y_{2}$ - Initial and final heights of the ball, measured in meters.

The final height of the ball is determined by the following formula:

$v_{2}^{2}+2\cdot g\cdot y_{2} = v_{1}^{2}+2\cdot g\cdot y_{1}$

$v_{1}^{2}-v_{2}^{2}+2\cdot g \cdot y_{1}=2\cdot g\cdot y_{2}$

$y_{2} = y_{1}+\frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g}$ (3)

If we know that $y_{1} = 30\,m$ , $v_{1} = 14\,\frac{m}{s}$ , $v_{2} = 0\,\frac{m}{s}$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height that the ball will reach is:

$y_{2} = 30\,m + \frac{\left(14\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$y_{2} = 39.993\,m$

The ball will reach a maximum height of 39.993 meters.

Given the absence of non-conservative forces, the ball exhibits a free fall. The time needed for the ball to reach its maximum height is computed from the following kinematic formula:

$t = \frac{v_{2}-v_{1}}{-g}$ (4)

If we know that $v_{1} = 14\,\frac{m}{s}$ , $v_{2} = 0\,\frac{m}{s}$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$t = \frac{0\,\frac{m}{s}-14\,\frac{m}{s} }{-9.807\,\frac{m}{s^{2}} }$

$t = 1.428\,s$

The ball will take 1.428 seconds to reach its maximum height.

6 0

3 years ago

What is the image point of (−2,5) after a translation right 3 units and up 4 units?

vfiekz [6]

Answer:

(1,9)

Step-by-step explanation:

(-2,5)

3 units right = x +3

4 units up = y + 4

(-2+3, 5+4)

(1,9

5 0

3 years ago