Help me plz will give BRIANLY

Which expression is represented by the diagram? o -6-(-1) o -6-1 o -5-(-1) o -5-1

cupoosta [38]

Option A:

–6 – (–1)

Solution:

Minus symbol are given in the diagram.

There are 6 minus symbol in the diagram.

From that expression 1 minus symbol is cancelled.

To find which expression represents this situation:

Option A: –6 – (–1)

Total number of minus = –6

From this cancel one minus: –6 – (–1)

–6 – (–1) = –6 + 1 = –5

So, remaining minus are –5.

This is the correct answer.

Option B: –6 –1

–6 –1 = –7

But note that remaining 5 minus are only there in the diagram.

So, this is not correct answer.

Option C: –5 – (–1)

Totally 6 minus in the problem. But here only 5 minus are given.

So, this is not correct answer.

Option D: –5 –1

Totally 6 minus in the problem. But here only 5 minus are given.

So, this is not correct answer.

Hence –6 – (–1) is the correct expression which represented the diagram.

8 0

3 years ago

Read 2 more answers

Bessie bought 10 gallons of Bush's Sweet Tea for a family reunion. How many liters did Bessie buy?

Liula [17]

Step-by-step explanation:

If 1 gallon = 3.785 litres

Then, 10 gallons = 37.85 litres

4 0

3 years ago

Find x, the unknown side. A) 4.6 C) 4.8 B) 5.4 D)3.5

MrMuchimi

Go ahead to do the trick

3 0

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