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miss Akunina [59]
3 years ago
7

A toy rocket launched straight up from the ground with an initial velocity of 80 ft/s returns to the ground after 5 s. The heigh

t of the rocket t seconds after launch is modeled by the function f(t)=−16t2+80t . What is the maximum height of the rocket, in feet?
Mathematics
2 answers:
eimsori [14]3 years ago
6 0
Maximum is 100
I just took the test, got it right 
mixas84 [53]3 years ago
5 0

Answer:

Maximum height oh the rocket was 100 feet.

Step-by-step explanation:

It has been given in the question that rocket follows the function f(t) = 80t-16t² where t is the time taken by rocket to reach maximum height and coming back to the ground.

Since initial velocity of the rocket is = 80 ft/sec

and total time taken by the rocket to come back to the ground is 5 sec

time taken by the rocket to reach the maximum height = 5/2 = 2.5 sec

Therefore from the given function

Maximum height achieved by the rocket = 80.(2.5) - 16.(2.5)²

= 200 - 100 = 100 ft.

 

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Solve for X<br> x/7 = -8<br> Please help!!!<br> Worth 20 points!!!<br> I need an answer ASAP!!!
Vikentia [17]

Answer:

x = -56

Step-by-step explanation:

x/7 = -8

Multiply each side by 7

x/7 *7 = -8*7

x = -56

8 0
3 years ago
Which of the following is true of the data represented by the box plot
zmey [24]

Answer:

The data is skewed to the bottom and contains an outlier.  

Step-by-step explanation:

1. Test for outlier

An outlier is a point that is more than 1.5IQR below Q1 or above Q3.  

IQR = Q3 - Q1 = 74 - 51 = 23

1.5 IQR  = 1.5 × 23 = 34.5

51 - 15 = 36 > 1.5IQR

The point at 15 is an outlier.

2. Test for normal distribution

The median is not in the middle of the box.  

Rather, it cuts the box into two unequal parts, so the data does not have a normal distribution.

3. Test for skewness

The longer part is to the left of the median, so the data is skewed left.

 

4 0
3 years ago
What is the amplitude of sin ?
sp2606 [1]

You haven't provided a graph or equation so I will tell the simplified meaning of amplitude instead.

Amplitude, is basically a distance from midline/baseline to the maximum or minimum point.

For sine function, can be written as:

\displaystyle \large{ y = A \sin(bx  -  c) + d}

  • A = amplitude
  • b = period = 2π/b
  • c = horizontal shift
  • d = vertical shift

I am not able to provide an attachment for an easy view but I will try my best!

We know that amplitude or A is a distance from baseline/midline to the max-min point.

Let's see the example of equation:

\displaystyle \large{y = 2 \sin x}

Refer to the equation above:

  • Amplitude = 2
  • b = 1 and therefore, period = 2π/1 = 2π
  • c = 0
  • d = 0

Thus, the baseline or midline is y = 0 or x-axis.

You can also plot the graph on desmos, y = 2sinx and you will see that the sine graph has max points at 2 and min points at = -2. They are amplitude.

So to conclude or say this:

If Amplitude = A from y = Asin(x), then the range of function will always be -A ≤ y ≤ A and have max points at A; min points at -A.

6 0
3 years ago
I will give brainliest!
Makovka662 [10]
I would think the answer is 3, sorry if i cant help. 
3 0
4 years ago
The Ace Novelty company produces two souvenirs: Type A and Type B. The number of Type A souvenirs, x, and the number of Type B s
Molodets [167]

Answer:

  500 type A; 3500 type B

Step-by-step explanation:

The method of Lagrange multipliers can solve this quickly. For objective function f(x, y) and constraint function g(x, y)=0 we can set the partial derivatives of the Lagrangian to zero to find the values of the variables at the extreme of interest.

These functions are ...

  f(x,y)=4x+2y\\g(x,y)=2x^2+y-4

The Lagrangian is ...

  \mathcal{L}(x,y,\lambda)=f(x,y)+\lambda g(x,y)\\\\\text{and the partial derivatives are ...}\\\\\dfrac{\partial \mathcal{L}}{\partial x}=\dfrac{\partial f}{\partial x}+\lambda\dfrac{\partial g}{\partial x}=4+\lambda (4x)=0\ \implies\ x=\dfrac{-1}{\lambda}\\\\\dfrac{\partial \mathcal{L}}{\partial y}=\dfrac{\partial f}{\partial y}+\lambda\dfrac{\partial g}{\partial y}=2+\lambda (1)=0\ \implies\ \lambda=-2

  \dfrac{\partial\mathcal{L}}{\partial\lambda}=\dfrac{\partial f}{\partial\lambda}+\lambda\dfrac{\partial g}{\partial\lambda}=0+2x^2+y-4=0\ \implies\ y=4-2x^2\\\\\text{We know $\lambda$, so we can find x and y:}\\\\x=\dfrac{-1}{-2}=0.5\\\\y=4-2\cdot 0.5^2=3.5

Since x and y are in thousands, maximum profit is to be had when the company produces ...

  500 Type A souvenirs, and 3500 Type B souvenirs

3 0
3 years ago
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