Plotting the data to get the line of best fit, it is:
y<span> = 0.087</span>x<span> + 0.587
substitute x=2030-1980=50,
y = $4.94 is the price in 2030</span>
Answer:
28
Step-by-step explanation:
Because 8 times 6 is 48 for the area so 6 and 8 multiply by two then combined is 28.
Answer:
he did half the diameter
Step-by-step explanation:
Answer:
a. A(x) = (1/2)x(9 -x^2)
b. x > 0 . . . or . . . 0 < x < 3 (see below)
c. A(2) = 5
d. x = √3; A(√3) = 3√3
Step-by-step explanation:
a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...
A(x) = (1/2)(x)(y)
A(x) = (1/2)(x)(9-x^2)
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b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).
On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).
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c. A(2) = (1/2)(2)(9 -2^2) = 5
The area is 5 when x=2.
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d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).
Answer:
22% = 0.22
Step-by-step explanation:
Lat'e express all the given numbers in decimal form so we understand how they are located on the number line.
We start with 2/11 , since the number we are looking for has to be larger than this \, and smaller than 1.42:
So we understand that we are looking for a number that can be placed between the lower boundary (0.181818...) and the upper boundary (1.42) on the number line. That is: we are looking for a number greater than 0.18181818... and less than 1.42.
Now let's look at each of the given options (also writing them in decimal form to facilitate the comparison with the lower and upper boundaries we just found):
This number is grater than the upper boundary given to us (1.42), it would be placed to the right of 1.42 on the number line. Therefore we discard it for not being in the requested interval (section) of the number line.
0.153 is already in decimal form, and clearly less than (thus to be placed on the left of) the lower boundary (0.181818...) of the requested interval. Therefore we discard it for not being in the requested interval (section) of the number line.
22% in decimal form is written as: . This number is greater than the lower boundary (0.181818...) and also less than the upper boundary (1.42). Therefore it is a choice that would make the sentence true.
which is clearly greater than the upper boundary of the interval, so we discard it.