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3 years ago

Choose the correct description of the graph of the inequality x − 3 greater than or equal to 5.

Mathematics

2 answers:

cricket20 [7]3 years ago

7 0

$x - 3 \geq 5 \\ \\ x \geq 5 + 3 \\ \\ x \geq 8$

The graph will be a shaded region to the right of the solid vertical line x =8

frez [133]3 years ago

4 0

Answer:

x ≥ 8

Step-by-step explanation:

Given: x - 3 ≥ 5

Now solve for x.

Add 3 on both sides, we get

x - 3 + 3 ≥ 5 + 3

x ≥ 8

To graph this inequality, draw a line line and mark 8 with a filled circle shade the region right of 8.

Herewith I have attached the graph.

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Determine the intercepts of the line.<br> y = 10x - 32<br> y-intercept: (<br> z-intercept: (

Rom4ik [11]

Answer:

see below

Step-by-step explanation:

y = 10x - 32

To find the x intercept set y =0 and solve for x

0 = 10x -32

Add 32 to each side

32 = 10x

Divide by 10

32/10=10x/10

3.2 = x

To find the y intercept set x =0 and solve for t

y = 10*0 -32

y= -32

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