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3 years ago

What is the area of a square with a side of 8x^7 y^3

1 answer:

4 0

Area of a square with side s is $\sf{s^2}$

In your question, the side or s is: $\sf{8x^7y^3}$

And so the area of a square with that side length would be:

$\sf{(8x^7y^3)^2}$

And using this formula: $\sf{(a^b)^c =a^{b\times c}}$

We get that the area is:

$\sf{(8x^7y^3)^2 = 8^2 x^{7\times 2} y^{3\times 2}}$

And simplifying that we get the final answer as:

$\huge{\boxed{\bf{Area=64x^{14}y^{6}}}}$

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Graph the inequality x<1

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3 years ago

Which function represents g(x), a refection of f(x)=1/2(3)^x across the y-axis

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Answer: g(x) = (1/2)3^-x reflection over y axis yields (-x,y)

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2 years ago

Point A is located at (-3, 9) and point B is located at (12,-10). Find the distance from point A to point B rounded to the neare

Nezavi [6.7K]

Answer:

Step-by-step explanation:

$\sqrt{(( - 10) - 9)^{2} + (12 - ( - 3)) ^{2} }$

$\sqrt{( - 19)^{2} + (15)^{2} }$

$\sqrt{361 + 225}$

$\sqrt{586}$

$24.2074$

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2 years ago

By what percent must we decrease the number A to obtain the number 4A/5?

Schach [20]

$\bf \cfrac{4A}{5}\implies \cfrac{4}{5}\cdot A\impliedby \textit{A is being reduced to four fifths} \\\\\\ \stackrel{\textit{let's convert four fifths to percentage}}{\cfrac{4}{5}\cdot 100\implies \cfrac{400}{5}\implies 80}$

so 4/5 of A is really 80% of A then.

since a whole A is 100% of A, 100% - 80% = 20%, A is being reduced by 20%, namely 1/5.

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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