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3 years ago

The length of a rectangular flower garden is 9 feet and it’s width is 5 feet how many feet of fencing will be needed

1 answer:

8 0

Use the simple perimeter formula.
P = 2L + 2W
P = 2(9) + 2(5)
P = 19 + 10
P = 29ft

Therefore 29ft of fencing will be needed.

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6.1 m A 85.4 sq m ⓅⓁⓈ ⒽⒺⓁⓅ

Mrac [35]

The teacher probably wants you to multiply cause area refers to that

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3 years ago

I am so stuck! :( Pls help me!!! I don't get the problem and I don't know how to solve it

Lelu [443]

Answer:

$\frac{n^{2}}{2a}$

Step-by-step explanation:

Just split it up first $\frac{9an^{3}}{18a^{2}n}=\frac{9}{18}\times \frac{a}{a^2} \times\frac{n^3}{n}$
Then 9 divided by 18 can be simplified into the fraction 1/2
What can go into $a$ and $a^2$ ? a itself right, so if you divide a by a you get 1, if you divide $a^2$ by a you get a
Same as the a fraction, if you divide $n^3$ by n you get $n^2$
$\frac{9an^{3}}{18a^{2}n}=\frac{9}{18}\times \frac{a}{a^2} \times\frac{n^3}{n}=\frac{1}{2}\times \frac{1}{a} \times\frac{n^2}{1}=\frac{n^{2}}{2a}$

4 0

3 years ago

Read 2 more answers

Please help me I'm stuck

AnnZ [28]

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6 0

3 years ago

HELP ILL GIVE BRAINLISTTT!

lions [1.4K]

Answer:

You can substitute

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(

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Step-by-step explanation:

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

4 0

3 years ago