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Iteru [2.4K]
3 years ago
12

Student finished 8 homework problems. If the ratio of the problems she finished to the problems she had left was 4:1 How many ho

me work problems did she have total
Mathematics
1 answer:
Readme [11.4K]3 years ago
4 0
Answer: 10 


Explanation: 

Problem finished : Problem left = 4 : 1

 4 : 1 ===> (x2) ===> 8 : 2 

Problem finished : Problem left = 8 : 2

Total homework = 8 + 2 + 10
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Answer:

123

Step-by-step explanation:

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Suppose an object is launched from ground level directly upward at 57.4 f/s Write a function to represent the object’s height ov
Semmy [17]

Answer: p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

Step-by-step explanation:

We can suppose that the only force acting on the object is the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration.

Then we can write:

a(t) = -32 ft/s^2

Where the negative sign is because this acceleration is downwards.

Now, to get the vertical velocity of the object, we need to integrate over time to get:

v(t) = (-32 ft/s^2)*t + v0

where t represents time in seconds and v0 is the constant of integration, and in this case, is the initial vertical velocity.

In this case, the initial velocity is 57.4 ft/s upwards, then the velocity equation is:

v(t) = (-32 ft/s^2)*t + 57.4 ft/s

To get the position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t + p0

Where p0 is the initial height of the object, as it was launched from the ground, then the initial position is p0 = 0ft.

then the position equation (that is the function that represents the height of the object as a function over time) is:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t

p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

3 0
3 years ago
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Answer:

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2 years ago
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<h2>>> Answer </h2>

___________

\:

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  • Step by step :

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x² = 3² + 4²

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3 years ago
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(-2m + 5n)²

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