On a test, the number of questions Pam answered correctly was

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GREYUIT [131]

The answer 0.1, 0.136, 0.652, 0.87

8 0

3 years ago

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Question 6 Multiple Choice Worth 1 points)

OleMash [197]

Answer:

Quantity A because it changes with gravitational pull.

Explanation:

Mass defines the constitutive matter of an object. It is immutable (does not change). Although it could be converted into energy, the matter is constant.

On the other hand, the weight of an object (gravitational force) depends on the acceleration due to gravity.

That is why an astronaut weighs about three times more on earth than on Mars. The gravitational acceleration on Mars is about one third that on earth.

5 0

3 years ago

Point A is located at (−2, −6), and D is located at (−6, 8). Find the coordinates of the point that lies halfway between A and D

Afina-wow [57]

Answer:

(-4,1)

Step-by-step explanation:

Point A is located at (−2, −6), and D is located at (−6, 8).

We need to find the midpoint of A and D

Mid point formula is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Point A is (-2,-6) that is (x1,y1)

Point D is (-6,8) that is (x2, y2)

plug in the values in the formula

$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

$(\frac{-2-6}{2}, \frac{-6+8}{2})$

$(\frac{-8}{2}, \frac{2}{2})$

Mid point is (-4, 1)

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