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Marat540 [252]
2 years ago
9

Answer part C. I want step by step on this problem.

Mathematics
1 answer:
elixir [45]2 years ago
4 0

Answer:

Rate = 58007956.5217

Step-by-step explanation:

Solving (c): The rate of change:

In this concept, the formula to use is:

Rate = \frac{V2 - V1}{t2 - t1}

Where

(t1,v1) = (0,62817000) ----- In 2004

(t2,v2) = (23,1397000000) ------ In 2026

So, we have:

Rate = \frac{1397000000 - 62817000}{23 - 0}

Rate = \frac{1334183000}{23}

Rate = 58007956.5217

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Please help ( 20 points )
VashaNatasha [74]

Answer:

1+pi/2, 5/2, square root of 12-1, 2.25. Let me know if this is correct!


Step-by-step explanation:


7 0
3 years ago
Which of the fractional factors will result in a product that increases when multiplied by the fraction in the image?
Delicious77 [7]

the fraction in the image is 2 1/2

let's try each choice

1) 2/3 x 5/2 = 10/6 = 5/3

2) 5/4 x 5/2 = 25/8

3) 3/3 x 2 1/2 = 2 1/2

4) 4/5 x 5/2 = 20/10 = 2

who is giving you these questions?

8 0
3 years ago
The force of gravity on Mars is different than on Earth. The function of the same situation on Mars would be represented by the
sweet-ann [11.9K]

Answer:

If thrown up with the same speed, the ball will go highest in Mars, and also it would take the ball longest to reach the maximum and as well to return to the ground.

Step-by-step explanation:

Keep in mind that the gravity on Mars; surface is less (about just 38%) of the acceleration of gravity on Earth's surface. Then when we use the kinematic formulas:

v=v_0+a\,*\,t\\y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2

the acceleration (which by the way is a negative number since acts opposite the initial velocity and displacement when we throw an object up on either planet.

Therefore, throwing the ball straight up makes the time for when the object stops going up and starts coming down (at the maximum height the object gets) the following:

v=v_0+a\,*\,t\\0=v_0-g\,*\,t\\t=\frac{v_0}{t}

When we use this to replace the 't" in the displacement formula, we et:

y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2\\y-y_0=v_0\,(\frac{v_0}{g} )-\frac{g}{2} \,(\frac{v_0}{g} )^2\\y-y_0=\frac{1}{2} \frac{v_0^2}{g}

This tells us that the smaller the value of "g", the highest the ball will go (g is in the denominator so a small value makes the quotient larger)

And we can also answer the question about time, since given the same initial velocity v_0 , the smaller the value of "g", the larger the value for the time to reach the maximum, and similarly to reach the ground when coming back down, since the acceleration is smaller (will take longer in Mars to cover the same distance)

3 0
2 years ago
If |x|+10=1, then what does x equal
Pie

Answer; there is literally no solution. it has to be a negative number.

6 0
3 years ago
Read 2 more answers
Helpp I’ll Mark you brainiest if your correct
patriot [66]
The answer is 586.17 cm^2
6 0
2 years ago
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