1: A bag contains 10 white golf balls and 6 striped golf balls. A golfer wants to add 122 golf balls to the bag. He wants the ra
1 answer:
Part A:
Given that a<span> bag contains 10 white golf balls and 6 striped golf balls, the ratio of white to striped golf balls is 10 : 6 = 5 : 3.
Given that </span>a<span> golfer wants to add 122 golf balls to the bag, the of balls each he should add is given by:
Therefore, the golfer shold add 76 white golf balls and 46 striped golf balls.
Part B.
You did not indicate the scale factor.
Assuming the scale factor is given by D(o, k) where k can be any number, then the cordinate of the </span><span>coordinates of the image of the quadrilateral VWXY are given by:
V'(6k, 2k), W(-2k, 4k), X(-3k, -2k), Y(3k,-5k)
That is you multiply the scale factor to the coordinates of the preimage to get the coordinates of the image.
</span>
Given that a<span> bag contains 10 white golf balls and 6 striped golf balls, the ratio of white to striped golf balls is 10 : 6 = 5 : 3.
Given that </span>a<span> golfer wants to add 122 golf balls to the bag, the of balls each he should add is given by:
Therefore, the golfer shold add 76 white golf balls and 46 striped golf balls.
Part B.
You did not indicate the scale factor.
Assuming the scale factor is given by D(o, k) where k can be any number, then the cordinate of the </span><span>coordinates of the image of the quadrilateral VWXY are given by:
V'(6k, 2k), W(-2k, 4k), X(-3k, -2k), Y(3k,-5k)
That is you multiply the scale factor to the coordinates of the preimage to get the coordinates of the image.
</span>
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The shape of the graph of quadratic functions is the shape of a parabola.
The steps for drawing a graph of the function f(x) = 2·(x + 4)² - 3 arranged in the correct order are;
- (a) Plot the vertex (-4, -3)
- (b) Apply the 2·x² pattern, from the vertex go right 1 up 2 and plot (-3, -1)
- (c) Apply the 2·x² pattern again, from the vertex go right 2 up 8 and plot (-2, 5)
- (d) Apply the 2·x² pattern a 3rd time, from the vertex go right 3 up 18 and plot (-1, 15)
- (e) Draw the axis of symmetry x = -4
- (f) Reflect the points over the axis of symmetry and plot (-5, -1), (-6, 5), and (-7, 15)
- (g) Draw a U shaped parabola.
Reasons:
The given function is; f(x) = 2·(x + 4)² - 3
The function is given in the vertex form; f(x) = a·(x - h)² + k
Therefore, the vertex, (h, k) = (-4, -3)
Step (a);
The vertex can be plotted on the graph
- Plot the vertex (-4, -3)
Step (b);
Given that the quadratic term is 2·x², the pattern that can be used for the points from the vertex is therefore, 2·x²
From the vertex (-4, -3) apply the 2·x² pattern by going to the right 1 unit and up 2 × 1² = 2 units to get the point (-4 + 1, -3 + 2) = (-3. -1)
- Apply the 2·x² pattern, from the vertex go right 1 up 2 and plot (-3, -1)
Step (c);
To get the next point, the 2·x² pattern is applied with x = 2, to the vertex to get; (-4 + 2, -3 + (2×2²)) = (-2, 5)
- Apply the 2·x² pattern again, from the vertex go right 2 up 8 and plot (-2, 5)
Step (d);
A third point on the graph relative to the vertex is obtained again by applying the 2·x² pattern again to the vertex with x = 3, to get;
(-4 + 3, -3 + (2 × 3²)) = (-1, 15)
- Apply the 2·x² pattern a 3rd time, from the vertex go right 3 up 18 and plot (-1, 15)
Step (e);
The axis of symmetry can be drawn with a vertical line passing through the vertex, which is the line, x = -4
The line x = -4 can be drawn on the graph next
- Draw the axis of symmetry x = -4
Step (f);
The points obtained relative to the vertex (-3, -1), (-2, 5), (-1, 15) can reflected about the axis of symmetry x = -4, to get;
(-3, -1) (-5, -1)
(-2, 5) (-7, 5)
(-1, 15) (-7, 15)
- Reflect the points over the axis of symmetry and plot (-5, -1), (-6, 5), and (-7, 15)
Step (g);
The parabola that is U shaped can be drawn from the points plotted in the steps above.
- Draw a U shaped parabola
Learn more about drawing the graph of a quadratic function here: