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tamaranim1 [39]
1 year ago
13

Cars A, B, and C are travelling at constant speeds. Select the car that travels the fastest. Choose 1 answer: Choose 1 answer: (

Choice A) A (Choice B) B Car B travels a distance of ddd kilometers in hhh hours, based on the equation 55h=d55h=d55, h, equals, d. (Choice C) C Car C travels 135135135 kilometers in 333 hours.
Mathematics
1 answer:
Gennadij [26K]1 year ago
6 0

Answer:

correct answer is B

Step-by-step explanation:

comparing the speed at which the three cars travel and also considering the fact that they travel at a constant speed without increase or decrease in speed or acceleration, it gives B the highest speed at 55h per day

thank you.

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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
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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
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3 years ago
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Vadim26 [7]

Answers/Step-by-step explanation:

A. LCM

B. Greatest Common Factor(GCF) shows the largest whole number, in this case patties and buns, would be a part of the whole that matches both numbers. Neil is unable to buy parts of packages because that not how most stores do business. Least common multiple(LCM) is the number that is both closest in value to the original number while being equal for all numbers. in that case, Neil is buying whole packages so it would work.

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Answer:

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Step-by-step explanation:

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