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3 years ago

Convert to a scientific notation 860,000,000,000 =

Mathematics

2 answers:

valentina_108 [34]3 years ago

4 0

$8\underbrace{60,000,000,000}_{\leftarrow11}=8.6\times10^{11}$

prohojiy [21]3 years ago

3 0

8.60 x 10 raised to the 11 power

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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 :)

5 0

3 years ago

) For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.) x = (No Response) For which values of x

grigory [225]

Answer:

Check below, please

Step-by-step explanation:

Step-by-step explanation:

1.For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.)

When the derivative of a function is equal to zero, then it occurs when we have either a local minimum or a local maximum point. So for our x-coordinates we can say

$f'(x)=0\: at \:x=2, and\: x=-2$

2. For which values of x is f '(x) positive?

Whenever we have

$f'(x)>0$

then function is increasing. Since if we could start tracing tangent lines over that graph, those tangent lines would point up.

$f'(x)>0 \:at [-4,-2) \:and\:(2, \infty)$

3. For which values of x is f '(x) negative?

On the other hand, every time the function is decreasing its derivative would be negative. The opposite case of the previous explanation. So

$f'(x)$

4.What do these values mean?

$f(x) \:is \:increasing\:when\:f'(x) >0\\\\f(x)\:is\:decreasing\:when f'(x)$

5.(b) For which values of x is f ''(x) zero?

In its inflection points, i.e. when the concavity of the curve changes. Since the function was not provided. There's no way to be precise, but roughly

at x=-4 and x=4

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3 years ago

What is the produce? -9x(5-2x)

Marta_Voda [28]

Answer:

279x

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

I need help on this question so if u actually can answer it correctly and show how you got it that would be helpful thx