Stage1-Math-3rdSem-AppActivity

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Stage 1: Relations and polynomial functions Application activity Part 1. Linear functions as mathematical models 1. When you get into a cab, there is an initial fixed rate of $8.00, plus the additional rate per kilometer of $4.50. determine the specific equation that can relate the payment fee with the kilometers traveled and make a graph to mark the function. How much would you play for a 25-kilometer journey? And, if you paid $48.50, how many kilometers did you traveled?

Payment fee: $8.00 Kilometers traveled: $4.50 Ride = 8+4.50x Ride of 25 km = 8+4.50(25) = 8+112.5 = 120.5 Ride of $48.50 = 4.5x + 8 = 48.5 =4.5x = 48.5 - 8 =4.5x = 40.5 = x = 40.5 / 4.5 =x=9


2. The owner of a hamburger place pays weekly $1175.00 to produce 35 hamburgers while producing 60 hamburgers costs him $1550.00. if each hamburger is $40.00 determine the following: a) The specific equation that can relate the expenses with the number of hamburgers made (cost effectiveness equation).

đ?‘š=

1550 − 1175 = 15 60 − 35

đ?‘Ś-1175 = 15(đ?‘Ľ − 35) đ?‘Ś = 15 đ?‘Ľ − 35 + 1175 đ?‘Ś = 15đ?‘Ľ − 525 + 1175 đ?’š = đ?&#x;?đ?&#x;“đ?’™ + đ?&#x;”đ?&#x;“đ?&#x;Ž b) The specific equation that can relate the income regarding the number of hamburgers sold (cost income equation).

đ?’Šđ?’?đ?’„đ?’?đ?’Žđ?’† = đ?&#x;’đ?&#x;Žđ?’™ c) The quantities of hamburgers that need to be made and sold in order that the owners won’t gain or lose money (balance point).

đ?‘–đ?‘›đ?‘?đ?‘œđ?‘šđ?‘’ = 40đ?‘Ľ đ?‘?đ?‘œđ?‘ đ?‘Ąđ?‘ = 40đ?‘Ľ đ?‘’đ?‘žđ?‘˘đ?‘–đ?‘™đ?‘–đ?‘?đ?‘&#x;đ?‘–đ?‘˘đ?‘š đ?‘?đ?‘œđ?‘–đ?‘›đ?‘Ą: 40đ?‘Ľ = 15đ?‘Ľ + 650 40đ?‘Ľ-15đ?‘Ľ = 650 25đ?‘Ľ = 650 đ?‘Ľ = 650/25 đ?‘Ľ = 26 You have to produce and sell 26 hamburgers. d) The utility function equation.

đ?’‘đ?’“đ?’?đ?’‡đ?’Šđ?’• = đ?&#x;?đ?&#x;“đ?’™-đ?&#x;”đ?&#x;“đ?&#x;Ž


e) The profit if they produce and sell 45 hamburgers.

đ?‘?đ?‘&#x;đ?‘œđ?‘“đ?‘–đ?‘Ą = 25đ?‘Ľ − 650 đ?‘?đ?‘&#x;đ?‘œđ?‘“đ?‘–đ?‘Ą = 25 45 − 650 đ?‘?đ?‘&#x;đ?‘œđ?‘“đ?‘–đ?‘Ą = 1125 − 650 đ?’‘đ?’“đ?’?đ?’‡đ?’Šđ?’• = đ?&#x;’đ?&#x;•đ?&#x;“ f) How many hamburgers are supposed to be produced and sold weekly to have a profit of $1100.00?

1100 = 25đ?‘Ľ − 650 25đ?‘Ľ = 1100 + 650 1750 đ?‘Ľ= 25 đ?’™ = đ?&#x;•đ?&#x;Ž 3. Study, which are the temperatures (In Celsius and Fahrenheit Degrees) of freezing, and boiling water? Once you have these, and based in the previous information, determine the following: a) The equation that associates the Fahrenheit degrees in accordance to the Celsius.

đ?’‡ = đ?&#x;?. đ?&#x;–đ?’™ + đ?&#x;‘đ?&#x;? b) Which is the equivalent of 20ÂşC to ÂşF?

đ?‘“ = 1.8đ?‘Ľ + 32 đ?‘“ = 1.8(20) + 32 đ?‘“ = 1.8(20) + 32 đ?’‡ = đ?&#x;”đ?&#x;– c) Which is the equivalent of 140ÂşF to ÂşC?

đ?‘“ = 1.8đ?‘Ľ + 32 đ?‘“-32/1.8 = đ?‘Ľ đ?‘Ľ = 140 − 32/1.8 đ?’™ = đ?&#x;”đ?&#x;Ž Âşđ?‘Ş


Part 2. The linear inequalities as mathematical model 1. Regarding problem 1 from Part 1, determine the following: a. The number of kilometers you need to travel so the fee dos no exceed $100.00. Travel= 8+4.50x Travel= 8+4.50x= $100 x= $100-8/4.50 x= 20.44 You need to travel 20.44km, so the fee does not exceed $100.00. b. How many kilometers would you need to travel so the fee is over $150.00? Travel= 8+4.50x Travel= 8+4.50x= $150 x= $150-8/4.50 x= 31.55 You need to travel 31.55 km, so the fee does not exceed $150.00. 2. According to problem 2 from Part 1, how many hamburgers are needed to be sold for the profit to be at least $2500.00? profit = 25x-650 = 2500 profit = x=2500+650/25 x= 126 There are 126 hamburgers needed to be sold for the profit be at least $2500.00. 3. According to problem from Part 1, which is the temperature in Degrees Celsius that is greater or equal than 104ÂşF? đ?’‡ = đ?&#x;?. đ?&#x;–đ?’™ + đ?&#x;‘đ?&#x;? 104ÂşF = 1.8x + 32 104-32/1.8 = x x= 40 40ÂşC is equal than 104ÂşF.


Part 3. The quadratic functions as mathematical models Instructions Before you start solving the following quadratic application problems, answer the next questions. a) What is the name of the following quadratic function graph?

b) What is the name of the only point in the graph where there is only a value of “x” for a given “y” value?

Intercept point point


c) Where does the graph opens to if “a” is positive? In this case, does the graph have a maximum or minimum value?

It opens upward of the graph, and thereby it has a minimum value. d) Where does the graph opens to if “a” is negative? In this case, does the graph have a maximum or minimum value?

It opens downward of the graph, and thereby it has a maximum value. e) Which is the formula to determine the value of “x” where it has the maximum or minimum value of the quadratic function? f) How can the corresponding maximum or minimum value be calculated?

Xv=-b/2a

Determining the values of “a”, “b” and “c” from the quadratic function: f(x)=ax2+bx+c, and then using them in the formula Xv=b/2a to obtain it. But to determine if its maximum or minimum value you have to identify if the parabola opens upward, you will be finding its minimum value, or if the parabola opens downward you will find its maximum value.


Problem 1: In Physics, in the topic of vertical shot upwards, there is a formula to calculate an object when is launched vertically upward at a certain initial speed over ground. Research this formula.

Having investigated the formula, what is the particular formula to calculate the height of an object launched vertically upward as an initial speed of 30 m/s?

We can use the next formula to calculate the height:

but as we can see, the formula needs the t value, so to obtain it we will need an extra formula: that formula is useful to obtain the time value. So, the process must be like: The velocity value is zero because that is the velocity reached at the highest point of any object when it is launched vertically upward.

The gravity value is negative because the object is launched against the gravity, that means that the force acts in the negative direction.

After obtaining the time value, we can see that we have already the values required to use the other formula:


Vo = 30 m/s V=0 Yo = 0 m t = 3.061 s g = 9.8 m/s2 Now let’s use the first formula to finally obtain the height:

Although, there is a formula that also we can use to obtain the height without the necessity of two formulas neither the time value:


And the operations process must be:

As we can see, we obtained the same result as in the past process where we used two formulas. This one was easier. If is thrown from a building of 13.5 m of height, calculate: a) The time that the object takes to reach his maximum height.

t = 3.061 s What element of the parabola represents the time of the object? b) The maximum height reached by the object over ground.

In this case we will need the other formula because now we have a change in the initial height = 13.5 m but now we won’t need to do the complete process because the time value is the same and we already obtained it.


In this case the object is found at an initial height of 13.5 meters because it is over a building with 13.5 meters of height.

What element of the parabola represents the maximum height reached?

The vertex


Problem 2: If you have 120 m to fence a land of rectangular form: a) Which are the measurements of the land for the area

Perimeter formula: P = L + L + L + L

2x + 2y = 120 2y = 120 – 2x 2x = 120 – 2y y = 60 – x or x = 60 – y

We know that the formula to obtain the area of a rectangle is: A = L x L So, using the previous value of “y” we will obtain the next equation: Y=60-x A=x(60-x) A=60x-x2


First, to obtain “x” value we will use the formula XV=-b/2a

60x -x

2

-x2 +60x +0 A B C Xv=-b/2a Xv=-(+60)/2(-1) Xv=-60/-2 Xv=30 This is the value given to “x” and the value for the vertex of the parabola. As we can see, 30 is positive, and that means the parabola opens downward, so the vertex value that we obtained (+30) is the maximum possible value. We can prove it graphing the equation, so you will obtain a parabola like the next one.


So, after obtained the value of “x”, we will be able to obtain the “y” value and maximum area of the rectangle. Y=60-x Y=60-(+30) Y=60-30 Y=30m

A=60x-x2 A=60(30)-(30)2 A=1800-900 A=900m2

The dimensions for the maximum area should be x = 30, y = 30 Because: 30 + 30 + 30 + 30 = 120 b) Which is the corresponding maximum area?

The maximum area is 900m2.


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