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Xelga [282]
3 years ago
6

Someone help please and make sure it’s right please :)

Mathematics
1 answer:
melamori03 [73]3 years ago
8 0
I'm going to go with 4 I hope I'm right if it's wrong I'm sorry
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C is correct. 
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The correct answer for the question that is being presented above is this one:

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For what value of x is the rational expression below equal to zero?<br> 2-X<br> 2 + x
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2

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Determine the slope and y-intercept from the following equation: 4x + y = -10

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3 years ago
On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John
insens350 [35]

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

     $H_{max}=\frac{v^2}{2g}$

     $H_{max} = \frac{(48)^2}{2 \times 12}$

               = 96 ft

b). Time he can stay in the air before hitting the ground is

   $T=\frac{2v}{g}$

  $T=\frac{2 \times 48}{12}$

     = 8 seconds

c).  Considering upward motion as positive direction.

     v = u + at

We find the time taken to reach the maximum height by taking v = 0.

     v = u + at

     0 = 16 + (12) t

     $t=\frac{16}{12}$

        $=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t =  $=\frac{4}{3} \ s$  , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$  feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

Therefore final velocity is = -16 ft/s

3 0
3 years ago
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