Is the square root of 62 Irrational or rational

Mathematics

2 answers:

worty [1.4K]3 years ago

7 0

62 is a rational number because it can be expressed as the quotient of two integers: 62÷ 1

Eddi Din [679]3 years ago

6 0

the square root of 62 is irrational

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Which of the following is a trinomial?

Sedaia [141]

Answer:

$A) c^2 + c + 6$

Step-by-step explanation:

We have been given four choices out of which we need to select the choice which is trinomial.

Trinomial means polynomial having three terms.

A) c2 + c + 6

It has 3 terms.

B) c2 − 16

It has 2 terms.

C) −8c

It has 1 term.

D) c3 + 4c2 − 12c + 7

It has 4 terms.

We see that only choice A) has three terms.

Hence correct choice is $A) c^2 + c + 6$

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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)