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3 years ago

I'll fan

Mathematics

2 answers:

Nezavi [6.7K]3 years ago

6 0

Remark: the x-intercepts are clearly (0, 0) and (20, 0), not (0, 0), (0, 20).

having the 2 x intercepts and the vertex we can draw the graph, check picture 1 attached.

Moreover, we can also determine the quadratic function as follows.

The quadratic function with roots a and b, is given by the equation:

f(x)=c(x-a)(x-b).

[ since f(a)=0, f(b)=0, f(x) must have one factor (x-a) and one (x-b)]

The x coordinates of the x intercepts are the roots of the function so with the roots given:

f(x)=c(x-0)(x-20)=cx(x-20), the vertex is at (10, 15) so

15=f(10)=c*10*(10-20)=c*10*(-10)=-100c
c=-15/100=-3/20

the quadratic function (expressed in factorized form) is

f(x)=-3/20x(x-20)

part A: the x intercept 0 is the point where the ball was hit, the x intercept 20 is the point where the ball fell back to the ground, 20 feet away from the kicker.

the function is increasing in the interval x∈(0, 10) and decreasing in x∈(10, 20)
this means that the height is increasing in the interval (0, 10) and decreasing as x goes through the interval (10,20)
the distance from the kicker is increasing during the whole interval (0, 20)

Part B: at x=15, f(x)=f(15)=-3/20*15(15-20)= $\frac{-3}{20}*15*(-5)= \frac{-3}{4}*3*(-5)= \frac{45}{4}$

at x=12, f(x)=f(12)=-3/20*12(12-20)= $\frac{-3}{20}*12*(-8)= \frac{-3}{5}*3*(-8)= \frac{72}{5}$

the average rate of change is the ratio of the differences in y to the differences in x : $\frac{y_1_5-y_1_2}{x_1_5-x_1_2}= \frac{\frac{45}{4}-\frac{72}{5}}{15-12}= -1.05$

so the average rate of change is -1.05. The minus, tells us that the height is decreasing in the following way: for 1 feet in distance, we are loosing 1.05 feet in height.

Tpy6a [65]3 years ago

6 0

Part A:The graph below shows the height of a kicked soccer ball f(x), in feet, depending on the distance from the kicker x, in feet

X Intercept=(0,0) it represents the time the ball was not kicked at all

X intercept=(16,0)

the feet away from the starting point is 16ft the location or distance the ball is away from the starting point once landed.

Maximum: is 10

intervals:it increased(0 to 8) decreased (8 to 16)

when the ball was 8 ft away from the kicker horizontally, it was 10 feet in the air then it landed an additional 8 feet away from the kicker that was 16 feet away one it landed after going a maximum of 10 feet.

Part B: m=(y2 - y1) / (x2 - x1)

(x1, y1) (x2, y2)

(8,10) (13, 7)

m = (7-10)/(13-8) = -3/5

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3 years ago

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

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a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

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