Ask question

Ask question

All categories

3 years ago

8

Victoria kicked a soccer ball laying on the ground. It was in the air for 6 seconds before it hit the ground. While the soccer b

all was in the air it reached a height of approximately 25 feet. Assuming that the soccer ball’s height (in feet) is a function of time (in seconds), what is the maximum in the context of this problem?
Group of answer choices

(6, 25)

(3, 12.5)

(3, 25)

(20, 3)

Flag this Question

1 answer:

aniked [119]3 years ago

7 0

Answer:3, 25

Step-by-step explanation:

The seconds should be the x-axis and the height is y axis. Also thinking about how a kicked ball makes an arch, the max is 25, and the mid point between 0 and 6 is 3.

You might be interested in

For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

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

or

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so:

$\theta\neq \pi$

having said this, we can now start solving the equation:

$tan(\frac{\theta}{2})=sin(\theta)$

we can start solving this equation by using the half angle formula, such a formula tells us the following:

$tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}$

so we can substitute it into our equation:

$\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)$

we can now multiply both sides of the equation by $sin(\theta)$

so we get:

$1-cos(\theta)=sin^{2}(\theta)$

we can use the pythagorean identity to rewrite $sin^{2}(\theta)$ in terms of cos:

$sin^{2}(\theta)=1-cos^{2}(\theta)$

so we get:

$1-cos(\theta)=1-cos^{2}(\theta)$

we can subtract a 1 from both sides of the equation so we end up with:

$-cos(\theta)=-cos^{2}(\theta)$

and we can now add $cos^{2}(\theta)$

to both sides of the equation so we get:

$cos^{2}(\theta)-cos(\theta)=0$

and we can solve this equation by factoring. We can factor $cos(\theta)$ to get:

$cos(\theta)(cos(\theta)-1)=0$

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

$cos(\theta)=0$

$\theta=cos^{-1}(0)$

$\theta={\frac{\pi}{2}, \frac{3\pi}{2}}$

Equation 2:

$cos(\theta)-1=0$

we add a 1 to both sides of the equation so we get:

$cos(\theta)=1$

$\theta=cos^{-1}(1)$

$\theta=0$

so we end up with three answers to this equation:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

7 0

3 years ago

what is the equation for the volume of a cylinder? I think is v=pi × r^2 * h just asking for clarification

lys-0071 [83]

Answer:

V=pi multiplied by r^2 times height

Step-by-step explanation:

3 0

3 years ago

What is the best estimation of the equation 7/8 -6/11

Elena L [17]

Answer:

125/88

Step-by-step explanation:

have a nice day

8 0

3 years ago

5. If position of object x = 3 sinΘ – 7 cosΘ then motion of object is bounded between position.

lesya692 [45]

9514 1404 393

Answer:

±√58 ≈ ±7.616

Step-by-step explanation:

The linear combination of sine and cosine functions will have an amplitude that is the root of the sum of the squares of the individual amplitudes.

|x| = √(3² +7²) = √58

The motion is bounded between positions ±√58.

_____

Here's a way to get to the relation used above.

The sine of the sum of angles is given by ...

sin(θ+c) = sin(θ)cos(c) +cos(θ)sin(c)

If this is multiplied by some amplitude A, then we have ...

A·sin(θ+c) = A·sin(θ)cos(c) +A·cos(θ)sin(c)

Comparing this to the given expression, we find ...

A·cos(c) = 3 and A·sin(c) = -7

We know that sin²+cos² = 1, so the sum of the squares of these values is ...

(A·cos(c))² +(A·sin(c))² = A²(cos(c)² +sin(c)²) = A²(1) = A²

That is, A² = (3)² +(-7)² = 9+49 = 58. This tells us the position function can be written as ...

x = A·sin(θ +c) . . . . for some angle c

x = (√58)sin(θ +c)

This has the bounds ±√58.

3 0

3 years ago

Please Help Me ASAP!!!

Arturiano [62]

Everything you need is in the picture attached.

4 0

3 years ago