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The first line is perpendicular, the second line is not perpendicular and not parallel , the third line is parallel to the above line and the forth line not parallel or perpendicular to the line.
<h3>How can the slope of the line be calculate?</h3>The slope of the line is been given as −3x+5y=−15
Then , hence the slope is
- Slope of the first line 5x + 3y = 15 after calculation is
the slope is
this means that the line is perpendicular to the given line
- Slope of the second line 3x + 5y = 15 after calculation is
This means that the line is not perpendicular and not parallel
- Slope of the Third line -3x + 5y = 15 can be calculated as
This means that the line is parallel to the above line.
- Slope of the first line 3x + 5y = 15 after calculation is
This means that the line is not parallel or perpendicular to the line
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The motion of the ball dropped from height is a free fall motion due to
gravitational acceleration.
<h3>Correct response;</h3>- The equation that models the height is; <u>y = -16·x²</u>
To select the equation that models the height, <em>y</em>, of the ball <em>x</em> seconds after
its dropped;
<h3>Solution:</h3>From the above table, we have that the first difference is not a constant
The second difference is = 48 - 16 = 32
Taking the second difference as a constant, we have the following
quadratic sequence;
y = a·x² + b·x + c
Where;
x = The time in seconds
y = The height after <em>x</em> seconds
- At x = 0, we have;
200 = a·0² + b × 0 + c
Therefore;
c = 0
- At<em> x</em> = 1, we have;
184 = a × 1² + b × 1 + 200
184 = a + b + 200
a + b = -16
a = -16 - b
- At <em>x</em> = 2, we have;
136 = a × 2² + b × 2 + 200
136 = 4·a + 2·b + 200
-64 = 4·a + 2×b
Therefore;
-64 = 4 × (-16 - b) + 2×b
-64 = -64 - 4·b + 2×b
b = 0
a + b = -16
Therefore;
a + 0 = -16
a = -16
The equation that models the height is; y = <u>-16·x²</u>
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