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liberstina [14]
2 years ago
10

You throw a softball into the air with an initial upward velocity of 38 ft/s and an initial height of 5 feet. Use the vertical m

otion model h=16t^2+vt+c to help you answer the questions below. (A sketch of the graph might help give you a visual of the scenario.) A) Create an equation using the given information. B) What is height of the softball after 2 seconds? C) What is the softball's maximum height? D) How long did it take the softball to reach it's max height? E) How long does it take for the softball to hit the ground after being thrown?
Mathematics
1 answer:
ddd [48]2 years ago
7 0
I’m so sorry I need points but I’m gonna manifest that someone smarter will see this and answer
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Given the system of constraints name all vertices. Then find the maximum value of the given objective
larisa86 [58]

Answer:

35

Step-by-step explanation:

The constraints are

The red line represents the function

y\leq \dfrac{1}{3}x+1

At y=0

0=\dfrac{1}{3}x+1\\\Rightarrow -1=\dfrac{1}{3}x\\\Rightarrow x=-3

At x=0

y=0+1\\\Rightarrow y=1

Two points are (-3,0),(0,1)

The blue line represents the function

5\geq y+x

at y=0

5=x

at x=0

y=5

Two points are (5,0),(0,5)

The other two constraints are x\geq 0, y\geq 0. So, the point has to be in the first quadrant

From the graph it can be seen there are two points where the function will be maximum let us check them.

(3,2)

7x-2y=7\times 3-2\times 2=17

(5,0)

7x-2y=7\times 5-2\times 0=35

So, the maximum value of the function is 35.

3 0
2 years ago
When Robert got a puppy from the shelter, it weigh 11 pounds. The puppy gained 40% of its original weight in the first month tha
Iteru [2.4K]
The puppy weighs 15.4 pounds, 40% of 11 is 4.4 so add that to the wight and its 15.4
3 0
3 years ago
The function f(t)=3,502.86(1.15)(t−12) models the amount of money in Gavin's investment account after t years. Gavin says this m
liraira [26]

Answer: Yes, because 3502.86 is multiplied by the exponent. i think, because i was stuck between that and the other yes, and the other yes was wrong, so im pretty sure its that.

Tell me if im right.

Step-by-step explanation:

5 0
3 years ago
Exhibit B: A restaurant has tracked the number of meals served at lunch over the last four weeks. The data shows little in terms
Mekhanik [1.2K]

Answer:

Option E is correct.

The expected number of meals expected to be served on Wednesday in week 5 = 74.2

Step-by-step Explanation:

We will use the data from the four weeks to obtain the fraction of total days that number of meals served at lunch on a Wednesday take and then subsequently check the expected number of meals served at lunch the next Wednesday.

Week

Day 1 2 3 4 | Total

Sunday 40 35 39 43 | 157

Monday 54 55 51 59 | 219

Tuesday 61 60 65 64 | 250

Wednesday 72 77 78 69 | 296

Thursday 89 80 81 79 | 329

Friday 91 90 99 95 | 375

Saturday 80 82 81 83 | 326

Total number of meals served at lunch over the 4 weeks = (157+219+250+296+329+375+326) = 1952

Total number of meals served at lunch on Wednesdays over the 4 weeks = 296

Fraction of total number of meals served at lunch over four weeks that were served on Wednesdays = (296/1952) = 0.1516393443

Total number of meals expected to be served in week 5 = 490

Number of meals expected to be served on Wednesday in week 5 = 0.1516393443 × 490 = 74.3

Checking the options,

74.3 ≈ 74.2

Hence, the expected number of meals expected to be served on Wednesday in week 5 = 74.2

Hope this Helps!!!

8 0
3 years ago
b) If parametric equations of a flow line are x = x(t), y = y(t), explain why these functions satisfy the differential equations
sineoko [7]

Answer:

The equation of the the flow line that passes through the point (x, y) = (−1, −1) is

In y + In x = 0 or in another form, xy = 1.

Step-by-step explanation:

The pathline equation for a vector field is given by F(x,y) = xî - yj

The velocity vector field for the streamline of the flow is given by

V(x, y) = (dx/dt)î + (dy/dt)j

From the question, it is given that

(dx/dt) = x

(dy/dt) = -y

Hence, the velocity vector field for the streamline of the flow in question is

V(x, y) = xî - yj

which coincides with the pathline vector field of the flow.

The only time the pathline and streamline vector field coincide and have the same equation is when the flow is a steady state flow.

That is, the properties of the fluid flowing isn't changing with time!

Hence, this flow is a steady state flow!

We're told to solve the differential equation.

(dx/dt) = x

(dy/dt) = -y

but

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = -y/x

(dy/y) = -(dx/x)

∫(dy/y) = -∫ (dx/x)

In y = - In x + c

where c is the constant of integration

In y + In x = c

In (xy) = c

Inserting the values of (x, y) given in the question,

In (-1 × -1) = c

In 1 = c

0 = c

c = 0

In y + In x = 0

In (yx) = 0

xy = e⁰ = 1

xy = 1

So, the equation of the the flow line that passes through the point (x, y) = (−1, −1) is

In y + In x = 0 or in another form, xy = 1

Hope this Helps!!!

4 0
3 years ago
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