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Sagot :
Answer:
Kangaroo's maximum height is 6 m and the kangaroo's jump is 28 m long
Step-by-step explanation:
Given :[tex]y = -0.03(x - 14)^2 + 6[/tex]
To Find : What is the kangaroo's maximum height? How long is the kangaroo's jump?
Solution:
[tex]y = -0.03(x - 14)^2 + 6[/tex]
x is the horizontal distance in meters
y is the vertical distance in meters for the height of the jump.
Substitute y = 0
[tex]0 = -0.03(x - 14)^2 + 6[/tex]
[tex]0.03(x - 14)^2 = 6[/tex]
[tex](x - 14)^2 = \frac{6}{0.03}[/tex]
[tex](x - 14)^2 =200[/tex]
[tex](x - 14) =\sqrt{200}[/tex]
[tex](x - 14) =14.142[/tex]
[tex]x =14.142+14[/tex]
[tex]x =28.142[/tex]
x≈ 28 m
Now the maximum height will be attained at mid point i.e. [tex]\frac{28}{2} =14[/tex]
Now substitute x= 14
[tex]y = -0.03(14 - 14)^2 + 6[/tex]
[tex]y = 6[/tex]
So, kangaroo's maximum height is 6 m and the kangaroo's jump is 28 m long
- From the vertex of the quadratic equation, we find that: The kangaroos maximum height is of 6 meters.
- From the roots of the equation, we find that: The kangaroo's jump is 28.14 meters long.
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Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
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Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
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The quadratic equation is:
[tex]y = -0.03(x - 14)^2 + 6[/tex]
Placing in standard form:
[tex]y = -0.03(x^2 - 28x + 196) + 6[/tex]
[tex]y = -0.03x^2 + 0.84x + 0.12[/tex]
Thus, it has coefficients [tex]a = -0.03, b = 0.84, c = 0.12[/tex]
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The kangaroo's maximum height is the y-value of the vertex, thus:
[tex]\Delta = b^2 - 4ac = (0.84)^2 - 4(-0.03)(0.12) = 0.72[/tex]
[tex]y_{v} = -\frac{\Delta}{4a} = -\frac{0.72}{4(-0.03)} = 6[/tex]
The kangaroos maximum height is of 6 meters.
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The length of the kangaroo's jump is the positive root. The roots are found at the values of x for which y = 0, thus, the solutions of the quadratic equation.
[tex]x_{1} = \frac{-0.84 + \sqrt{0.72}}{2(-0.03)} = -0.14[/tex]
[tex]x_{2} = \frac{-0.84 - \sqrt{0.72}}{2(-0.03)} = 28.14[/tex]
The kangaroo's jump is 28.14 meters long.
A similar question is given at https://brainly.com/question/16858635
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Use A Table To Find Two Consecutive Integers Between Which The Solution Lies. 4k-13=12 24.1=15q-32.5
