Line segment AB has a length of 5 units. It is translated 3 units to the right on a coordinate

Please Help!!!

marishachu [46]

The second one, and you as well ☺️

8 0

3 years ago

If X=2 y=-1 z=3 Find the value of a. 2y3(cube) z2(square)

qaws [65]

Step-by-step explanation:

2y³z²

2(-1)³(3)²

=2(-1)(9)

=-18

6 0

3 years ago

Consider two functions: g(x)=−x2−6x and the quadratic function f(x) shown in the table.

S_A_V [24]

Answer:

Hence average rate of change is greater for f(x) in (0,3).

Yes. f(x) is greater than g(x) in the interval (0,3)

Yes. g(3) <f(3)

Step-by-step explanation:

To find average rate of change of f and g in (0,3)

f(3)-f(0)/3 = (9-0)/3 =3

g(3) = -9-18 = -27

g(0) = 0

Rate of change in(0,3) of g(x) = -27/3 =-9

Hence average rate of change is greater for f(x) in (0,3)

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Both f and g have intercepts of y as 0

Hence both are equal.

3) Yes. f(x) is greater than g(x) in the interval (0,3)

Because g(x) <0 for all x in (0,3) while f(x) >0

4) g(3)= -9-18=-27 <f(3)

5 0

3 years ago

Read 2 more answers

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

3 years ago

Math help please help me with this page

White raven [17]

Please, post just one problem at a time, or (if you post more than one), indicate which one you want to focus on first. Even more important, please do whatever you can to get started on each problem; I'm sure you know at least some basics.

What does "intersecting" mean? Look it up if you're not sure.

Then what would "two intersecting lines" look like? Draw the graph, or at least explain in words what the graph would look like.

5 0

2 years ago