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1 year ago

6

A tennis court is surrounded by a fence so that the distance from each boundary of the tennis court to the fence is the same, x

feet. If the tennis court is 78 feet long and 36 feet wide, which expression gives the area of the entire surface inside the fence?

1 answer:

nevsk [136]1 year ago

7 0

Given the information, we have the following:

since the distance from each boundary of the tennis court to the fence is the same (x feet), then we have:

therefore, the expression that gives the are of the entire surface inside the fence is A=(36+2x)(78+2x)

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Read 2 more answers

How to show these 2 problems are inverses

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$\bf \begin{cases}
f(x)=\sqrt[3]{7x-2}\\\\
g(x)=\cfrac{x^3+2}{7}
\end{cases}\\\\
-----------------------------\\\\
now
\\\\
f[\ g(x)\ ]\implies f\left[ \frac{x^3+2}{7} \right]\implies \sqrt[3]{7\left[ \frac{x^3+2}{7} \right]-2}\implies \sqrt[3]{x^3+2-2}
\\\\\\
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-----------------------------\\\\
or
\\\\
g[\ f(x)\ ]\implies g\left[\sqrt[3]{7x-2}\right]\implies \cfrac{\left[\sqrt[3]{7x-2}\right]^3+2}{7}
\\\\\\
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thus f[ g(x) ] = x indeed, or g[ f(x) ] =x, thus they're indeed inverse of each other

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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

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Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

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