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3 years ago

Solve the equation sin theta = 0.3 to the nearest tenth. Use the restrictions 90degrees < theta < 180degrees.

1 answer:

5 0

We first do:
∅ = sin⁻¹(0.3)
∅ = 17.5°

To go into the second quadrant, we add 90°.
∅ = 17.5 + 90
= 107.5°
The answer is A.

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Use a system of linear equations with two variables and two equations to solve. A number is 9 more than another number. Twice

Troyanec [42]

9514 1404 393

Answer:

(x, y) = (3, 12)

Step-by-step explanation:

Using the given variable definitions, we can write the equations ...

y = x + 9

2(x +y) = 30

Solving the second equation for y, we have ...

y = 15 -x

Substituting this into the first equation gives ...

15 -x = x +9

6 = 2x . . . . . . . . . add x-9

3 = x . . . . . . . . . . divide by 2

y = 3+9 = 12 . . . . substitute into the first equation

The numbers are (x, y) = (3, 12).

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2 years ago

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2 years ago

1. Find the value of x in the equation 2N x 20 15 31 X O A. 1 B. 16 OD. 25 ext Review Answer selections

Solnce55 [7]

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3 years ago

A 2.2 kg ball strikes a wall with a velocity of 7.4 m/s to the left. The ball bounces off with a velocity of 6.2 m/s to the righ

Naya [18.7K]

Answer:

The constant force exerted on the ball by the wall is 119.68 N.

Step-by-step explanation:

Consider the provided information.

It is given that the mass of the ball is m = 2.2 kg

The initial velocity of the ball towards left is 7.4 m/s

So the momentum of the ball when it strikes is = $2.2\times 7.4=16.28$

The final velocity of the ball is -6.2 m/s

So the momentum of the ball when it strikes back is = $2.2\times -6.2=-13.64$

Thus change in moment is: $16.28-(-13.64)=29.92$

The duration of force exerted on the ball t = 0.25 s

Therefore, the constant force exerted on the ball by the wall is:

$\frac{29.92}{0.25}=119.68$

Hence, the constant force exerted on the ball by the wall is 119.68 N.

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3 years ago

Help with this math problem, please!!

VMariaS [17]

Answer:

Step-by-step explanation:

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3 years ago