hectictar

Let's divide the pentagon into shapes like this:

By SSS congruency, the two triangles at the top are congruent, and so they have the same area.

area of one of the triangles  =  (1/2)(base)(height)  =  (1/2)(8)(6)  =  24  sq inches

area of rectangle  =  (length)(width)  =  (12)(8)  =  96  sq inches

area of pentagon  =  area of rectangle + 2(area of one of the triangles)

area of pentagon  =  96 sq inches + 2(24 sq inches)

area of pentagon  =  96 sq inches + 48 sq inches

area of pentagon  =  144 sq inches

By the distance formula...

distance between  (2, 5)  and  (-6, y)   =   \(\sqrt{(-6-2)^2+(y-5)^2}\)

The problem tells us that the distance between  (2, 5)  and  (-6, y)  is  10 units, so.....

\(\sqrt{(-6-2)^2+(y-5)^2}=10\)

                                                          Square both sides of the equation.

(-6 - 2)² + (y - 5)²  =  100

                                                          Multiply out each exponent on the left side.

( -8 )² + (y - 5)(y - 5)  =  100

64 + y² - 10y + 25  =  100

                                                          Combine  64  and  25  to get  89

y² - 10y + 89  =  100

                                                          Subtract  100  from both sides of the equation.

y² - 10y - 11  =  0

                                                          Factor the left side.

(y - 11)(y + 1)  =  0

                                                          Set each factor equal to zero and solve for  y .

y - 11  =   0       or       y + 1  =  0

   y  =  11                      y  =  -1

Since  y > 0  ,  the solution must be  y = 11  

Happy (belated) first anniversary!!!! Hope to see you stick around for many more!! 

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The probability that a student chosen at random has a cat and a dog   =   9 / 25

4 sin²x   =   4 cos x  +  5

                                                   By the Pythagorean identity,   sin²x  =  1 - cos²x    so....

4( 1 - cos²x )   =   4 cos x  +  5

                                                   Distribute the  4  to the terms in parenthesees.

4  -  4 cos²x   =   4 cos x  +  5

                                                   Add   4 cos² x   to both sides of the equation.

4   =   4 cos²x  +  4 cos x  +  5

                                                   Subtract  4  from both sides of the equation.

0   =   4 cos²x  +  4 cos x  +  1

                                                   Split   4 cos x   into two terms that we can use to factor by grouping.

0   =   4 cos²x  +  2 cos x  +  2 cos x  +  1

0   =   2 cos x ( 2 cos x  +  1 )  +  1( 2 cos x  +  1 )

0   =   ( 2 cos x  +  1 )( 2 cos x  +  1 )

0   =   ( 2 cos x  +  1 )²

                                      Take the square root of both sides.

0   =   2 cos x  +  1

                                      Subtract  1  from both sides.

-1   =   2 cos x

                                      Divide both sides by  2 .

-1/2   =   cos x

The angle that has a cosine of   -1/2   in the interval  [0, pi)   is   2 pi / 3

x  =  2 pi / 3

Let  V  be the volume of the balloon.

Let  r₁  be the radius of the sphere.

Let  r₂  be the radius of the hemisphere.

\(V=\frac43\pi (r_{\small1})^3\\~\\ \frac3{4\pi}\cdot V=\frac3{4\pi}\cdot\frac43\pi (r_{\small1})^3\\~\\ \frac{3V}{4\pi}=(r_{\small1})^3\\~\\ \sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{(r_{\small1})^3}\\~\\ \sqrt[3]{\frac{3V}{4\pi}}=r_{\small1}\)                 Solve this equation for  r₁

\(V=\frac12\cdot\frac43\pi (r_{\small2})^3\\~\\ V=\frac23\pi (r_{\small2})^3\\~\\ \frac3{2\pi}\cdot V=\frac3{2\pi}\cdot\frac23\pi (r_{\small2})^3\\~\\ \frac{3V}{2\pi}=(r_{\small2})^3\\~\\ \sqrt[3]{\frac{3V}{2\pi}}=\sqrt[3]{(r_{\small2})^3}\\~\\ \sqrt[3]{\frac{3V}{2\pi}}=r_{\small2} \)                  Solve this equation for  r₂ .

The ratio of  r₁  to  r₂  can be expressed in the form   \(\sqrt[3]{a}\)   for some real number  a .

\(\dfrac{r_{\small1}}{r_{\small2}}=\sqrt[3]{a} \\~\\ \dfrac{\sqrt[3]{\frac{3V}{4\pi}}}{\sqrt[3]{\frac{3V}{2\pi}}}=\sqrt[3]{a}\\~\\ \dfrac{\frac{3V}{4\pi}}{\frac{3V}{2\pi}}=a\\~\\ \frac{3V}{4\pi}\cdot\frac{2\pi}{3V}=a\\~\\ \frac14\cdot\frac21=a\\~\\ \frac24=a \\~\\ \frac12=a\)      Plug in the equivalent expressions of  r₁  and  r₂  and solve for  a .

Also.....

50 m : 30 m   \(=\frac{50\,\text{m}}{30\,\text{m}}=\frac{5\,\text m}{3\,\text m}=\frac53=\)   5 : 3

6 ft  to  6 yd   \(=\frac{6\text{ ft}}{6\text{ yd}}=\frac{1\text{ ft}}{1\text{ yd}}=\frac{1\text{ ft}}{3\text{ ft}}=\frac13=\)   1  to  3

\(\frac{120\text{ ft}}{13.36\text{ cm}}=\frac{120\text{ ft}\cdot\frac{1\text{ ft}}{1\text{ ft}}}{13.36\text{ cm}} =\frac{120\text{ ft}\cdot\frac{30.48\text{ cm}}{1\text{ ft}}}{13.36\text{ cm}} =\frac{3657.6\text{ cm}}{13.36\text{ cm}}=\frac{3657.6}{13.36}=\frac{100}{100}\cdot\frac{3657.6}{13.36}= \frac{365760}{1336}=\frac{45720}{167}\)

volume of pyramid  =  (1/3)(area of base)(height)

Let right triangle  ABC  be the pyramid base and  HD  be the height.

Let's find the edge length of the cube.

volume of cube  =  (edge length)³

1  =  (edge length)³

1  =  edge length

Let's find the area of right triangle ABC .

area of triangle ABC  =  (1/2)(AB)(BC)

area of triangle ABC  =  (1/2)(1)(1)

area of triangle ABC  =  1/2

HD  is an edge of the cube so

HD  =  1

volume of pyramid  =  (1/3)(area of triangle ABC)(HD)

volume of pyramid  =  (1/3)(1/2)(1)

volume of pyramid  =  1/6      cubic units