In the diagram a set of parallel lines are cut by a transversal.

Mathematics

2 answers:

kifflom [539]3 years ago

3 0

Answer:

C. Angle 3 and 4 are vertical angles

saw5 [17]3 years ago

3 0

Answer:

D. Angles 4 and 5 are alternate interior angles.

Step-by-step explanation:

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Which BEST represents the height of a home?

Sliva [168]

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3 years ago

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Find three consecutive integers whose sum is 57????

Setler79 [48]

Answer:18 19 20

Step-by-step explanation:"Consecutive" means that the integers will follow each other in value, for example: 1, 2, 3 or 4, 5, 6. Also, no decimals are needed here because "integers" are whole, counting numbers. Here is the set up: Let x= the first integer Then X+1= 2nd consecutive integer and x+2= 3rd .

Suppose that x=1 x+1= 1+1=2 and x+2=1+2=3 However, you need specific consecutive numbers whose sum is 57. Remember that sum means to add:

x+ (x+1) + (x+2) = 57 Addition of all 3 consecutive numbers Now solve for x

and substitute into each part to come up with the three integers:

3x + 3= 57 3x=54 x=54/3 = 18 x=18, x+1= 18+1=19 x+2=18+2=20

Check your answer: 18+19+20=57 57=57 Check

7 0

3 years ago

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 $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$