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Free_Kalibri [48]
2 years ago
12

Describe how to use the vertical line test and what it can tell us

Mathematics
1 answer:
sashaice [31]2 years ago
4 0

Answer:

To use the vertical line test, take a ruler or other straight edge and draw a line parallel to the y-axis for any chosen value of x. If the vertical line you drew intersects the graph more than once for any value of x then the graph is not the graph of a function.

Step-by-step explanation:

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Simply 2 x b x b x b x 4
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2×4=8
b×b×b= b^3
so putting them together would be
8×b^3
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What is the pre-image of vertex A' if the image shown on
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Answer: B. (-6, 8).

Step-by-step explanation:

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As an estimation we are told £3 is €4. Convert £34.50 to euros.
Andre45 [30]

Answer:

46 euros

Step-by-step explanation:

£30 = 40 euros

4.50 = £3 and £1.50

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1.50 = 2 euros because you half the 4 euros because half of 3 is 1.50

hope this helps if not sorry  

3 0
3 years ago
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An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
2 years ago
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