HELP PLEASE!!

Mathematics

1 answer:

pogonyaev2 years ago

3 0

Answer:

You need to have both the opposite and adjacent side of the right angled triangle. Tan = Opp/Adj

Step-by-step explanation:

Tangents are pretty much everywhere. A good example is maybe a slide at the playground. You’d need the tangent to figure out the angle of the slide and how it is.

The radius from the center of a circle to the point of tangency point shows that it would be perpendicular to the tangent line considering that anything with a radius is circular and the tangent line is… a line, making it impossible to be parallel.

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Which of the following is equivalent to the equation below

NNADVOKAT [17]

1- A = B = m²-16

2- D = 4x²-3 proof: (8x³-6x)/2x =2x(4x²-3)/2x simplify to get the answer

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What is the image of (4,2) ?

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An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object

Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

$1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2 + \frac{1}{2} at^2\\3) v = v_0 + at$

What do we already know?

$v_0 = 29.4 ms^-1$
The launch is straight up
$a = -9.81 ms^-2$ this is the gravitational acceleration $g$

$s_0 = 0 m$ , since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, $v = 0 ms^-2$

$v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m$

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

$s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2$

this is the equation of the whole trajectory that object makes.

and if you solve this by making $s = 0$ , you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)