for the school play adult tickets cost 4$ and children tickets cost 2$ natalie is working at the ticket counter and just sold 20
1 answer:
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<em>x-> points scored by John's team</em>
<em>
</em>
<em>=> 20x=1400</em>
<em>=><u> x=70</u></em>
<em>
<em>=> 20x=1400</em>
<em>=><u> x=70</u></em>
Given: <em>The height y of a ball (in feet) is given by the function </em><em>y=-1/12x^2+2x+4 </em><em>and x is the horizontal distance traveled by the ball.</em>
Part A:<em> </em><em>How high is the ball when it leaves the child's hand?</em>
Right after the ball leaves the child's hand, it has travelled 0 feet horizontally. Horizontal distance is represented by x, so we could say that x = 0.
Plug in 0 for our equation and solve for y, the height.
Part B & C: <em>How high is the ball at its maximum height?
</em>
What we basically want to do is find the vertex of the function.
There are multiple ways to do this. You could graph it or make a table, but this method is not efficient.
The method I am going to go over right now is putting the equation in vertex form.
Move the constant to the left side.
Factor out the x² coefficient.
Find out which number to add to create a perfect square trinomial.
(Half of 24 is 12, 12 squared is 144. We have to add 144/-12 (which is -12) to each side so that we end up with 144 inside the parentheses on the right side)
Factor the perfect square trinomial and simplify the right side.
Isolate y on the left side.
And now we are in vertex form.
Vertex form is defined as y = a(x-h)² + k with vertex (h, k).
In this case, our vertex is (12, 16).
You could've also taken the shortcut that for any quadratic f(x) = ax² + bx + c, the vertex (h, k) is (-b/2a, f(h)). That's basically a summation of this method which you can use if your teacher has taught it to you.
Part D & E: <em>What is the horizontal distance travelled by the ball when it hits the ground?</em>
When the ball hits the ground, y is going to be 0, since y is the ball's height.
There are many ways to solve a quadratic...split the middle, complete the square, and the quadratic formula.
<u>
</u><u>Solving by splitting the midlde</u>
If your quadratic has fractions, this is not a good option.
<u>
</u><u>Solving by completing the square</u>
Move the constant over the right side.
Divide by the x² coefficient.
(Dividing by -1/12 is the same as multiplying by its reciprocal, -12.)
Simplify the right side.
Halve the x coefficient, square it, and then add it to each side.
(Half of -24 is -12, and -12 squared is 144.)
Factor the perfect square trinomial.
Take the square root of each side.
192 = 8 × 8 × 3, so we can simplify √192 to 8√3.
Add 12 to each side and we get our answer.
Our function does not apply when x or y is less than 0, of course.
12-8√3 is negative, so this cannot be our answer.
So, the ball had travelled 12+8√3 feet at the time when it hit the ground.
<u>Solving with the quadratic formula</u>
For any equation ax² + bx + c = 0, the solution for x is
.
Our equation, y=-1/12x^2+2x+4, has a = -1/12, b=2, and c=4.
Let's plug these values into the quadratic formula.
Dividing by a fraction is the same as multiplying by its reciprocal...
Of course, we only want the positive value, 12+8√3.
Revisiting Part B & C:
Since parabolae are symmetrical, if you know two values of x for some value of y (like the x-intercepts we just found in part B) then you can find the average between them to find what the x value of the vertex is, then plug that in to find the y value of the vertex (the height we want)
The average between 12+8√3 and 12-8√3 is 12. Plug that in and we get 16!
Part A:<em> </em><em>How high is the ball when it leaves the child's hand?</em>
Right after the ball leaves the child's hand, it has travelled 0 feet horizontally. Horizontal distance is represented by x, so we could say that x = 0.
Plug in 0 for our equation and solve for y, the height.
Part B & C: <em>How high is the ball at its maximum height?
</em>
What we basically want to do is find the vertex of the function.
There are multiple ways to do this. You could graph it or make a table, but this method is not efficient.
The method I am going to go over right now is putting the equation in vertex form.
Move the constant to the left side.
Factor out the x² coefficient.
Find out which number to add to create a perfect square trinomial.
(Half of 24 is 12, 12 squared is 144. We have to add 144/-12 (which is -12) to each side so that we end up with 144 inside the parentheses on the right side)
Factor the perfect square trinomial and simplify the right side.
Isolate y on the left side.
And now we are in vertex form.
Vertex form is defined as y = a(x-h)² + k with vertex (h, k).
In this case, our vertex is (12, 16).
You could've also taken the shortcut that for any quadratic f(x) = ax² + bx + c, the vertex (h, k) is (-b/2a, f(h)). That's basically a summation of this method which you can use if your teacher has taught it to you.
Part D & E: <em>What is the horizontal distance travelled by the ball when it hits the ground?</em>
When the ball hits the ground, y is going to be 0, since y is the ball's height.
There are many ways to solve a quadratic...split the middle, complete the square, and the quadratic formula.
<u>
</u><u>Solving by splitting the midlde</u>
If your quadratic has fractions, this is not a good option.
<u>
</u><u>Solving by completing the square</u>
Move the constant over the right side.
Divide by the x² coefficient.
(Dividing by -1/12 is the same as multiplying by its reciprocal, -12.)
Simplify the right side.
Halve the x coefficient, square it, and then add it to each side.
(Half of -24 is -12, and -12 squared is 144.)
Factor the perfect square trinomial.
Take the square root of each side.
192 = 8 × 8 × 3, so we can simplify √192 to 8√3.
Add 12 to each side and we get our answer.
Our function does not apply when x or y is less than 0, of course.
12-8√3 is negative, so this cannot be our answer.
So, the ball had travelled 12+8√3 feet at the time when it hit the ground.
<u>Solving with the quadratic formula</u>
For any equation ax² + bx + c = 0, the solution for x is
Our equation, y=-1/12x^2+2x+4, has a = -1/12, b=2, and c=4.
Let's plug these values into the quadratic formula.
Dividing by a fraction is the same as multiplying by its reciprocal...
Of course, we only want the positive value, 12+8√3.
Revisiting Part B & C:
Since parabolae are symmetrical, if you know two values of x for some value of y (like the x-intercepts we just found in part B) then you can find the average between them to find what the x value of the vertex is, then plug that in to find the y value of the vertex (the height we want)
The average between 12+8√3 and 12-8√3 is 12. Plug that in and we get 16!