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kolbaska11 [484]
1 year ago
5

In 1980 approximately 4,825 million metric tons of carbon dioxide emissions were recorded for the United States. That number ros

e to approximately 6,000 million metric tons in the year 2005. Here you have measurements of carbon dioxide emissions for two moments in time. If you treat this information as two ordered pairs (x, y), you can use those two points to create a linear equation that helps you make predictions about the future of carbon dioxide emissions!A) Organize the measurements into ordered pairs. B) Find the slope,C) Set up an equation in point-slope form,D) Show the equation in slope-intercept form,E) Predict emissions for the year 2020,
Mathematics
1 answer:
Oxana [17]1 year ago
8 0

ANSWER and EXPLANATION

A) To organize the measurements in ordered pairs implies that we want to put them in the form:

(x_1,y_1);(x_2,y_2)

Therefore, the measurements in ordered pairs are:

\begin{gathered} (1980,4825) \\ (2005,6000) \end{gathered}

Note: 4825 and 6000 are in millions (10⁶) of metric tons

B) To find the slope, apply the formula:

m=\frac{y_2-y_1}{x_2-x_1}

Therefore, the slope is:

\begin{gathered} m=\frac{6000-4825}{2005-1980} \\ m=\frac{1175}{25} \\ m=47\text{ million metric tons per year} \end{gathered}

C) To find the in point-slope form, we apply the formula:

y-y_1=m(x-x_1)_{}

Therefore, we have:

y-4825=47(x-1980)

Note: the unit is in million metric tons

D) To show the equation in point-slope form, we have to put it in the form:

y=mx+b

To do that, simplify the point-slope form of the equation:

\begin{gathered} y-4825=47(x-1980) \\ y=47x-93060+4825 \\ y=47x-88235 \end{gathered}

E) To predict the emissions for the year 2020, substitute 2020 for x in the equation above:

\begin{gathered} y=47(2020)-88235 \\ y=94940-88235 \\ y=6705\text{ million metric tons} \end{gathered}

That is the prediction for the year 2020.

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$15 x 12Mths = $180

2 x 20 = $40 carlo gets back

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x = 12

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Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

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Because this is the same as to calculate the limit from the left and right side, of g(x).

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