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3 years ago

12

The area of a racquetball court is 800 square feet.The length of a standard court is twice as its width. What are the dimensions

of a standard racquetball court?

1 answer:

zubka84 [21]3 years ago

3 0

Given - area of a racquet ball court is 800 square feet.

length is the twice of the width

find out thedimensions of the racquet ball court.

To proof -

Formula

area of a rectangle = length×breadth

as given inthe question the length of court is twice of its width .

let assume width be w

than length become 2w

put in the formula

w×2w = 800 ft²

2w² = 800 ft²

w = $\sqrt{400}$ ft

w = 20 ft

Than the length become 40 ft.

Hence proved

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Right triangle ABC and its image, triangle A'B'C' are shown in the image attached.

AlladinOne [14]

Answer:

See explanation

Step-by-step explanation:

Triangle ABC ha vertices at: A(-3,6), B(0,-4) and (2,6).

Let us apply 90 degrees clockwise about the origin twice to obtain 180 degrees clockwise rotation.

We apply the 90 degrees clockwise rotation rule.

$(x,y)\to (y,-x)$

$\implies A(-3,6)\to (6,3)$

$\implies B(0,4)\to (4,0)$

$\implies C(2,6)\to (6,-2)$

We apply the 90 degrees clockwise rotation rule again on the resulting points:

$\implies (6,3)\to A''(3,-6)$

$\implies (4,0)\to B''(0,-4)$

$\implies (6,-2)\to C''(-2,-6)$

Let us now apply 90 degrees counterclockwise rotation about the origin twice to obtain 180 degrees counterclockwise rotation.

We apply the 90 degrees counterclockwise rotation rule.

$(x,y)\to (-y,x)$

$\implies A(-3,6)\to (-6,-3)$

$\implies B(0,4)\to (-4,0)$

$\implies C(2,6)\to (-6,2)$

We apply the 90 degrees counterclockwise rotation rule again on the resulting points:

$\implies (-6,-3)\to A''(3,-6)$

$\implies (-4,0)\to B''(0,-4)$

$\implies (-6,2)\to C''(-2,-6)$

We can see that A''(3,-6), B''(0,-4) and C''(-2,-6) is the same for both the 180 degrees clockwise and counterclockwise rotations.

7 0

3 years ago

How do i put this in slope-intercept form. passes through 2,-2 and 4,-1

enot [183]

Answer:

$\large\boxed{y=\dfrac{1}{2}x-3}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

<em>m</em><em> - slope</em>

<em>b</em><em> - y-intercept</em>

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have two points (2, -2) and (4, -1).

Substitute:

$m=\dfrac{-1-(-2)}{4-2}=\dfrac{-1+2}{2}=\dfrac{1}{2}$

Put the value of a slope and the coordinates of the point (2, -2) to the equation of a line:

$-2=\dfrac{1}{2}(2)+b$

$-2=1+b$ <em>subtract 1 from both sides</em>

$-3=b\to b=-3$

Finally:

$y=\dfrac{1}{2}x-3$

5 0

3 years ago

Please help me . What is x equal to?

777dan777 [17]

Answer:

1. x = 2

2. x = 6

3. x = 6

4. x = 1

Step-by-step explanation: i think thats it

8 0

3 years ago

Read 2 more answers

........................................

Sergio [31]

What is your Question...

3 0

3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago