Answer:
Step-by-step explanation:
f(x)=2(x-2)²-5
f(x)=2(x²-4x+4)-5
f(x)=2x²-8x+3
Answer:
<h2>12x - 20y = -9</h2>
Step-by-step explanation:
The standard form of an equation of a line:

We have

Convert
<em>multiply both sides by 10</em>
<em>subtract 24x from both sides</em>
<em>change the signs</em>
<em>divide both sides by 2</em>

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,

Split up the interval
![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
into

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]](https://tex.z-dn.net/?f=%5Bx_0%2Cx_1%5D%5Ccup%5Bx_1%2Cx_2%5D%5Ccup%5Ccdots%5Ccup%5Bx_%7Bn-2%7D%2Cx_%7Bn-1%7D%5D%5Ccup%5Bx_%7Bn-1%7D%2Cx_n%5D)
where

and

. Each subinterval has measure (width)

.
Now denote the left- and right-endpoint approximations by

and

, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are

. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints,

.
So, you have


Now let

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,

Factoring out

and regrouping the terms, you have

which is equivalent to

and is the average of

and

.
So the trapezoidal approximation for your problem should be
Answer:
m ≥ –6
Step-by-step explanation:
m+5 ≥ –1
subtract 5 from both sides
m ≥ -6
The answer is .1
........