hectictar

\(\phantom{=\,}(-8)^{-\frac23}\\ \)

\( =\,\frac{1}{(-8)^{\frac23}}\)          because      \(x^{-a}\,=\,\frac{1}{x^a}\)

\( =\,\frac{1}{(-8)^{\frac23}}\cdot\frac{(-8)^\frac13}{(-8)^\frac13}\)

\( =\,\frac{(-8)^{\frac13}}{(-8)^{\frac23+\frac13}}\)          because     \(x^a\cdot x^b\,=\,x^{a+b}\)

\( =\,\frac{(-8)^{\frac13}}{(-8)^{\frac33}}\)

\( =\,\frac{(-8)^{\frac13}}{(-8)^{1}}\)

\( =\,\frac{-2}{-8}\)

\( =\,\frac14\)

\(\frac{7\sqrt{(2a)^2+(1)^2}-4a^2-1}{\sqrt{1+4a^2}+3}\,=\,2 \)

Multiply both sides of the equation by   \(\sqrt{1+4a^2}+3\)   and note that  \(\sqrt{1+4a^2}+3\neq0\\~\\ \sqrt{1+4a^2}\neq-3\)

But there isn't a value of  a  that makes that true.

\(7\sqrt{(2a)^2+(1)^2}-4a^2-1\,=2(\sqrt{1+4a^2}+3)\\~\\ 7\sqrt{4a^2+1}-4a^2-1\,=\,2\sqrt{1+4a^2}+6\\~\\ 7\sqrt{4a^2+1}\,=\,2\sqrt{4a^2+1}+7+4a^2\\~\\ 7\sqrt{4a^2+1}-2\sqrt{4a^2+1}\,=\,7+4a^2\\~\\ 5\sqrt{4a^2+1}\,=\,7+4a^2\\~\\ (5\sqrt{4a^2+1})^2\,=\,(7+4a^2)^2\\~\\ 25(4a^2+1)\,=\,(7+4a^2)(7+4a^2)\\~\\ 100a^2+25\,=\,49+56a^2+16a^4\\~\\ -16a^4+44a^2-24\,=\,0\\~\\ -16(a^2)^2+44(a^2)-24\,=\,0\)

Let  a²  =  u  , and let's substitute   u  in for  a² .

\(-16u^2+44u-24\,=\,0\\~\\ -4u^2+11u-6\,=\,0\\~\\ -4u^2+8u+3u-6\,=\,0\\~\\ -4u(u-2)+3(u-2)\,=\,0\\~\\ (u-2)(-4u+3)\,=\,0\\~\\ u-2=0\qquad\text{or}\qquad-4u+3=0\\~\\ \phantom{--}u=2\qquad\text{or}\qquad u=\frac34\)

Now substitute  a²  back in for  u .

\(\begin{array}{ccc} a^2=2&\qquad\text{or}\qquad&a^2=\frac34\\~\\ a=\pm\sqrt2&\qquad\text{or}\qquad&a=\pm\sqrt{\frac34}\\~\\ &&a=\pm\frac{\sqrt3}{2}\\~\\ a=\sqrt2\qquad\text{or}\qquad a=-\sqrt2&\text{or}&a=\frac{\sqrt3}{2}\qquad\text{or}\qquad a=-\frac{\sqrt3}{2} \end{array}\)

The greatest value of  a  that satisfies the equation is  \(\sqrt2\)  .

I think that the  l 's  are  absolute value marks

a · b  =  20      Divide both sides of this equation by  b  to solve for  a .

a  =  20 / b

b · c  =  12      Divide both sides of this equation by  b  to solve for  c .

c  =  12 / b

a + b + c  =  12

                                    Substitute  20/b  in for  a ,  and substitute  12/b  in for  c .

\(\frac{20}{b}\) + b + \(\frac{12}{b}\)  =  12

                                   Multiply through by  b .

20 + b² + 12  =  12b

                                   Combine  20  and  12  to get  32 .

b² + 32  =  12b

                                   Subtract  12b  from both sides.

b² - 12b + 32  =  0

                                   Factor the left side.

(b - 4)(b - 8)  =  0

                                   Set each factor equal to zero and solve for  b .

b - 4  =  0       or       b - 8  =  0

   b  =  4         or         b  =  8

Now let's use these values of  b  to find  a  and  c .

If  b = 4  , then   a  =  20 / b  =  20 / 4  =  5

If  b = 4 ,  then   c  =  12 / b  =  12 / 4  =  3

So a solution is:   a  = 5 ,   b = 4 ,   and   c = 3

If  b = 8  , then   a  =  20 / b  =  20 / 8  =  2.5

If  b = 8 ,  then   c  =  12 / b  =  12 / 8  =  1.5

So a solution is:   a  = 2.5 ,   b = 8 ,   and   c = 1.5

The painting is in the center of the wall, so the distance between the end of the wall and the corresponding edge of the painting is the same on both sides.

The distance between the end of the wall and the corresponding edge of the painting is  x  ft.

The wall is 24 ft wide, and the painting is 10 ft wide.

So it kinda looks like this:

Here is how I'd find x :

x + 10 + x  =  24

2x + 10  =  24

2x  =  14

x  =  7

The only option that  x = 7  is a solution to is  B . 

By the Pythagorean identity....

\(\sin^2\theta+\cos^2\theta\,=\,1\\~\\ \frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}\,=\,\frac{1}{\cos^2\theta}\\~\\ \tan^2\theta+1\,=\,\sec^2\theta\)   Let's divide both sides of this equation by cos²θ, and simplify.

Now plug in  -135  for  sec θ   and solve the equation for  tan θ .

\( \tan^2\theta+1\,=\,(-135)^2\\~\\ \tan^2\theta+1\,=\,18225\\~\\ \tan^2\theta\,=\,18225-1\\~\\ \tan^2\theta\,=\,18224 \)

Since  cot θ  is positive,  tan θ  is positive. So take the positive square root of both sides.

\(\tan\theta\,=\,\sqrt{18224}\\~\\ \tan\theta\,=\,4\sqrt{1139}\)

And...

\(\frac{\sec\theta}{\tan\theta}\,=\,\sec\theta\div\tan\theta\,=\,\frac{1}{\cos\theta}\div\frac{\sin\theta}{\cos\theta}\,=\,\frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}\,=\,\frac{1}{\sin\theta}\,=\,\csc\theta\)

So....

\(\csc\theta\,=\,\frac{\sec\theta}{\tan\theta}\,=\,\frac{-135}{4\sqrt{1139}}\,=\,\frac{-135\sqrt{1139}}{4556}\\~\\ \csc\theta\,=\,\frac{-135\sqrt{1139}}{4556}\)

-9a - 7x - 9c - 9b - 7y

                                          Numbers can be added in any order, so we can rearrange the expression.

=   -9a - 9b - 9c - 7x - 7y

                                          Factor  -9  out of the first three terms and  -7  out of the last two terms.

=  -9(a + b + c) - 7(x + y)

                                          Substitute  -1  in for  a + b + c  and substitute  7  in for  x + y .

=  -9( -1 ) - 7( 7 )

=  9 - 49

=  -40

Ohh......I should have read that more carefully!!!

I tried to do these the way the instructions say to do it...

(a)      \(\frac{12\text{ in}}{1\text{ ft}}=\frac{x\text{ in}}{19\text{ ft}}\\~\\ \frac{12\text{ in}\cdot19\text{ ft}}{1\text{ ft}}=x\text{ in}\\~\\ 228\text{ in}=x\text{ in}\)

So..    19 ft  =  228 in

(b)      \(\frac{1\text{ yd}}{3\text{ ft}}=\frac{x\text{ yd}}{19\text{ ft}}\\~\\ \frac{1\text{ yd}\cdot19\text{ ft}}{3\text{ ft}}=x\text{ yd}\\~\\ \frac{19}{3}\text{ yd}=x\text{ yd}\\~\\ 6.3\text{ yd}\approx x\text{ yd} \)

So..    19 ft  ≈  6.3 yd

(18)

Here is one way to look at it....

This kite is made up of four right triangles:

The total area of the kite is the sum of the areas of these four triangles.

area of kite  =  area of red △ + area of blue △ + area of green △ + area of purple △

Now we need to find the area of each triangle.

area of a triangle  =  (1/2)(length of base)(height)

area of red triangle  =  (1/2)(7)(3)

area of blue triangle  =  (1/2)(7)(3)

area of green triangle  =  (1/2)(3.5)(3)

area of purple triangle  =  (1/2)(3.5)(3)

Now we can find area of kite.

area of kite  =  (1/2)(7)(3) + (1/2)(7)(3) + (1/2)(3.5)(3) + (1/2)(3.5)(3)

area of kite  =  (7)(3) + (3.5)(3)

area of kite  =  3(7 + 3.5)

area of kite  =  3(10.5)

area of kite  =  31.5        That is in ft²