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3 years ago

Write an equation of the line that passes through a pair of points (2,3),(4,-2)

Mathematics

1 answer:

Ne4ueva [31]3 years ago

3 0

Answer:

y=-5/2x+13

Step-by-step explanation:

m=(-2)-3/4-2

m=-5/2

y-y=m(x-x1)

y-3=-5/2(x-2)

y-3=-5/2x+10

+3 +3

y=-5/2x+13

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The ratios in a equivalent ratio table are 3:12, 4:16, and 5:20. If the ratio is 10, what is the second number? Justify your rea

anzhelika [568]

Answer:

Step-by-step explanation:

If you look at the ratios you can see the pattern of the first number being multiplied by 4 everytime. Therefore you must multiply 10 by 4.

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3 years ago

Thank you:)))))))))))))

Anastasy [175]

Your welcomeeeeeeeeeee

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3 years ago

Which of the following is a trinomial?

Masja [62]

Although you didn't provide any answer choices:

A trinomial is an expression with exactly three terms that can't be combined an example would be this

x^2 - 3x + 1

None of those terms can be combined

x^2 + 3 - 4 has three terms but it isn't considered a trinomial because you can simplify it to x^2 - 1 which is known as a binomial because it has 2 terms in it's simplified form.

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4 years ago

Read 2 more answers

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)