What is the image of the point (7,-5) after a rotation of 270° counterclockwise
2 answers:
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Answer:
Step-by-step explanation:
This quadratic models the parabolic motion equation
where h is the height of the object after the motion occurs, a is the pull of gravity, v(i) is the initial vertical velocity of the object, and h(i) is the height from which the object was initially launched. If we are looking for when the ball hits the ground, h = 0, because the height of an object on the ground has no height at all (or 0 height). To find the time the object was in the air, factor the quadratic and solve for the values of t.
Two factors of 4 that add up to be 3 and multiply to be 4 are -1 and 4. So t=-1 and t = 4. Since we know that time can never be negative, then t = 4 seconds.
Let z = 3x+2. Then, by substitution, your equation is
.. z^2 +7z -8 = 0
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This equation can be factored as
.. (z +8)(z -1) = 0
.. x = -8 or +1
Then
.. 3x +2 = -8 or +1
.. 3x = -10 or -1
.. x = -10/3 or -1/3
.. z^2 +7z -8 = 0
_____
This equation can be factored as
.. (z +8)(z -1) = 0
.. x = -8 or +1
Then
.. 3x +2 = -8 or +1
.. 3x = -10 or -1
.. x = -10/3 or -1/3