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OLEGan [10]
2 years ago
12

A ball is thrown from an initial height of

Mathematics
1 answer:
Eduardwww [97]2 years ago
8 0

Answer:

Step-by-step explanation:

If we are looking for the times that the ball was 11 meters off the ground, we sub in 11 for the height on the left and solve for t:

11=-5t^2+30t+1 and

0=-5t^2+30t-10 and factor this however it is you are factoring in class to solve for t to get

t = .35 seconds and t = 5.6 seconds

Because the ball reaches this point in its way up and then passes it again on its way down, the ball will have 2 times at this height.

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Answer:

72

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Which of the following graphs shows the solution to the system of equations y=5x-1 and y= x+3
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3 years ago
Solve the system 6x -2y+z= -2 2x+ 3y - 3z =11 x+ 6y=31
weqwewe [10]

Answer:

x = 1

y = 5

z = 2

Step-by-step explanation:

System of equations:

6x - 2y + z = -2

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x + 6y = 31

Isolate one variable in any of the equations:

x + 6y = 31

x = 31 - 6y

Plug in this value for x in another equation:

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Plug in these values in the remaining equation:

2(31 - 6y) + 3y - 3(-188 + 38y) = 11

62 - 12y + 3y + 564 - 114y = 11

626 - 12y + 3y - 114y = 11

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Plug in value of y into our other answers to solve for x and z:

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Check your work:

6x - 2y + z = -2

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Correct!

*Note there are several ways to solve for these types of problems. I used substitution*

8 0
3 years ago
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Add the left over 2, so 680 + 2

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I hope this helps!

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>
6 0
3 years ago
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