(4w-3)^2 what would be the answer to that?? please explain, i’m so confused.
2 answers:
<h3>Answer: 16w^2 - 24w + 9</h3>
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Explanation:
(4w-3)^2 is the same as (4w-3)(4w-3) because squaring something is the same as multiplying it by itself.
Another example would be something like 7^2 = 7*7
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So we're asked to expand out (4w-3)(4w-3)
Let's make x = 4w-3 and replace that first copy of '4w-3' with 'x'
We get this expression: x(4w-3)
Distribute the x through to get...
x(4w-3)
x*4w + x*(-3)
4wx - 3x
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Now replace the x with (4w-3) and apply distribution twice more
4wx - 3x
4w(4w-3) - 3(4w-3)
16w^2 - 12w - 12w + 9
16w^2 - 24w + 9 which is the answer
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You can use the FOIL rule as an alternative method to get this answer. The box method is a visual tool to keep all the terms organized. See the attached image below for what I mean by the box method. The four terms inside the boxes represent the four terms from 16w^2 - 12w - 12w + 9, which was the second to last step we got shown above.
Yet another alternative you could use is the perfect square formula which is
(a-b)^2 = a^2 - 2ab + b^2
where in this case, a = 4w and b = 3.
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An object that is in motion as a projectile follows a path or trajectory of a parabola
The function and values are;
- a) The equation of the quadratic function is;
- b) The maximum height of the ball is approximately <u>7.334 m</u>
- c) Horizontal distance at maximum height <u>18.8 meters</u>
<u />
Reason:
a) Known parameters are;
Let f(x) = a·x² + b·x + c represent the equation of the parabola modelling the path of the ball, we have;
Points on the path of the parabola = (0, 0), (35, 1.5), 37, 0)
Plugging the values gives;
0 = a·0² + b·0 + c
Therefore, c = 0
1.5 = 35²·a + 35·b
0 = 37²·a + 37·b
Solving gives;
a = -3/140, b = 111/140
The equation of the quadratic function is therefore;
b) The maximum height is given by the vertex of the parabola
The x-coordinate at the vertex is the point
Which gives;
The maximum height is therefore;
The maximum height of the ball is approximately 7.334 m
c) The distance the ball has travelled to horizontally is given by half of the range, <em>R</em> as follows;
The range of the motion, R = 37 meters
Therefore;
The distance the ball has travelled horizontally to reach the maximum height horizontally <u>18.5 meters</u>
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