Whats 6/7 divided by 1/2 i need it as a fraction not decimal

HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

AlekseyPX

Answer:

The correct answer if F

Step-by-step explanation:

Hope you have a great day or afternoon :D

8 0

3 years ago

Read 2 more answers

Find the coordinates of a point that divides the directed line segment PQ in the ratio 5:3. A) (2, 2) B) (4, 1) C) (–6, 6) D) (4

Yuri [45]

Answer:

The answer is explained below

Step-by-step explanation:

The question is not complete we need point P and point Q.

let us assume P is at (3,1) and Q is at (-2,4)

To find the coordinate of the point that divides a line segment PQ with point P at $(x_1,y_1)$ and point Q at $(x_2,y_2)$ in the proportion a:b, we use the formula:

$x-coordinate:\\\frac{a}{a+b}(x_2-x_1)+x_1 \\\\While \ for\ y-coordinate:\\\frac{a}{a+b}(y_2-y_1)+y_1$

line segment PQ is divided in the ratio 5:3 let us assume P is at (3,1) and Q is at (-2,4). Therefore:

$x-coordinate:\\\frac{5}{5+3}(-2-3)+3 \\\\While \ for\ y-coordinate:\\\frac{5}{5+3}(4-1)+1$

4 0

3 years ago

A sunflower is 15 inches tall and grows 2 inches a week. How many weeks will it take for the

Zarrin [17]

Answer:

4 weeks

Step-by-step explanation:

15 + 2x = 23

2x = 8

x = 4

8 0

2 years ago

Tessellation that use more than one type of regular polygon are called regular tessellations

NARA [144]

Answer:

FALSE

Step-by-step explanation:

A tessellation refest to a shape that is repeated over and over again covering a plane without any gaps or overlaps. The statement is false given that regular tessellations use only one polygon. Semi-regular tessellations are created with more than one type of regular polygon.

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

4 0

3 years ago