PLEASE ANYONE HELP !!!!

Mathematics

1 answer:

podryga [215]1 year ago

8 0

Answer:

28 should do it

Step-by-step explanation:

35+35+82= 152

180-152=28

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Someone solve and show your work. Thanks! I’m giving brainliest.

Dmitry [639]

Lol i was working on this earlier. First get rid of the parenthesis by distributing the powers to the other powers.

27/4x(6)y(8)
the () are for powers

3 0

3 years ago

Find the least common denominators of these two rational experssions -1/y and 2/5y

algol13

The least common denominator is 5y, and can be obtained by multiplying
-1/y by 5/5 to get -5/5y.

If we add the numbers using the least common denominator, we get:
-5/5y + 2/5y = -3/5y

6 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)