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Mathematics

2 answers:

nlexa [21]3 years ago

4 0

0.89 is the surface area

expeople1 [14]3 years ago

3 0

Answer:

The surface area is 80.9.

Step-by-step explanation:

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Mr. Richard's science class is conducting a variety of experiments with projectiles and falling objects. Two groups of students

UNO [17]

Answer:

1. Group C; 2. Group B; 3. Group D; 4. Group A

Step-by-step explanation:

These equations are in the form

$y=-4.9t^2+v_0t+h_0\\\text{or}\\y=-16t^2+v_0t+h_0$ , where v₀ is the initial velocity and h₀ is the initial height.

The first equation has no value for v₀ and a value of 19 for h₀. This means there is no velocity, so the ball is dropped, and since the initial height is 19, it is dropped from 19 meters. This makes it group C.

The second equation has a value of 50 for v₀ and no value for h₀. This means the initial velocity is 50 and there is no initial height. This makes it group B.

The third equation has no value for v₀ and a value of 50 for h₀. This means there is no initial velocity, so the ball is being dropped, and the initial height is 50. This makes it group D.

The fourth equation has a value of 19 for v₀ and no value for h₀. This means the initial velocity is 19 and there is no initial height. This makes it group A.

7 0

3 years ago

Read 2 more answers

M44b

viktelen [127]

Answer:

84.85 feet or 60 $\sqrt2$ feet.

Step-by-step explanation:

Given that:

Distance of paths between each pair of consecutive bases is 60 ft.

To find:

Distance that the ball needs to travel if it is thrown directly from home plate to second base?

Solution:

Actually the given problem can be mapped to a right angled triangle in which base and height are given to be equal to 60 ft (because each pair of consecutive bases are 60 ft apart), Plus, they are at right angles to each other.

So, kindly refer to the image attached.

AB and BC are the distance of paths between consecutive bases.

And we have to find AC.

We can use Pythagorean theorem to find out the Hypotenuse of the triangle.

According to Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AC^{2} = AB^{2} + BC^{2} \\\Rightarrow AC^{2} = 60^{2} + 60^{2} \\\Rightarrow AC = \sqrt{2 \times 60^2}\\\Rightarrow AC = \bold{60\sqrt{2}\ ft = 84.85\ ft}$