How do I answer this question❓ And tell me how to show my work! <br> Thank you!! 20 points ⭐️

Mila [183]

Answer:

m(u) = -0.25(u +2)² +1 or -0.25u² -u

Step-by-step explanation:

The equation is fairly easily written in vertex form, as the vertex point is on a grid line intersection at (-2, 1). The parabola opens downward, so the scale factor is negative.

The vertical change from the vertex is only a fraction of a unit when u differs from the vertex by 1. It is 1 unit when u differs from the vertex by 2, so the magnitude of the vertical scale factor is 1/2² = 1/4.

Our equation will be of the form ...

m(u) = (vertical scale factor)(u - (horizontal vertex location))² + (vertical vertex location)

For this graph, the equation is ...

m(u) = -0.25(u +2)² +1

or, simplifying, we get ...

m(u) = -0.25u² -u

8 0

3 years ago

A 1:35 scale model of a fishing hut is 17 cm tall, 18 cm wide, and 19.7 cm long. What are the dimensions of the actual ice fishi

konstantin123 [22]

Multiply each dimension by 35:-

17*35 = 595 cm

18*35 = 630 cm

19.7 * 35 = 689.5 cm

in meters the dimensions are L = 6.895 m, W = 6.3 m and H = 5.95 m

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