hectictar

Let's call the unlabeled side of the large triangle  " x ".

And let's call the height of that triangle  " h ". 

By Pythagorean theorem, we can make the equation

x² + y²   =   (16 + 4)²

                                    Subtract  y²  from both sides of the equation.

x²   =   (16 + 4)² - y²

Again by the Pythagorean theorem, we can make the equation

16² + h²   =   x²

                            Subtract  16²  from both sides of the equation.

h²   =   x² - 16²

                                             Substitute   (16 + 4)² - y²   in for   x²

h²   =   (16 + 4)² - y² - 16²

Again by the Pythagorean theorem, we can make the equation

4² + h²   =   y²               Substitute   (16 + 4)² - y² - 16²   in for  h²

4² + (16 + 4)² - y² - 16²   =   y²      Now we can solve for  y .

                                                     Simplify the left side of the equation.

4² + 20² - y² - 16²   =   y²

16 + 400 - y² - 256   =   y²

160 - y²  =  y²

                          Add  y²  to both sides of the equation.

160   =   2y²

                          Divide both sides by  2

80   =   y²

                          Take the positive square root of both sides.

√80  =  y

y  =  √[2*2 * 2*2 * 5]

y  =  4√5

Given that the point  (8, 8)  is on the graph of  \(y=\frac 14f\left(\frac 12x\right)\) , there is one point that must be on the graph of  y = f(x) . What is the sum of the coordinates of that point?

\(8=\frac 14f\left(\frac 12(8)\right)\\~\\ 8=\frac 14f\left(4\right)\\~\\ 4\cdot8=f\left(4\right)\\~\\ 32=f\left(4\right)\\~\\ f\left(4\right)=32\)

So  (4, 32)  must be a point on the graph of  y = f(x) .

4 + 32  =  36

Let  QR  be the base of the triangle.

Let's draw a height from  QR  to  P .

Now we have a right triangle.

sin( angle )  =  opposite / hypotenuse

sin 20°  =  h / 9

                            Multiply both sides of the equation by  9 .

9 sin 20°  =  h

h  =  9 sin 20°

Now we know the base and the height, so we can find the area.

area  =  (1/2)(base)(height)

area  =  (1/2)(14)(9 sin 20°)   square cm

area  =  63 sin 20°   square cm

area  ≈  21.5   square cm

If we took off  1  inch from each stick, we'd have sticks of length

7, 14, and 16

For these to be the lengths of the sides of a triangle, the sum of the smallest two sides must be greater than the third side.

7 + 14 > 16

21 > 16          this is true, so  7, 14, and 16 can be the sides of a triangle.

So.....we want to know the smallest integer  n  such that....

(8 - n) + (15 - n)   is NOT greater than   17 - n

(8 - n) + (15 - n)  ≤  17 - n

23 - 2n  ≤  17 - n

23  ≤  17 + n

6  ≤  n

n  ≥  6

We want to know the smallest integer  n  such that  n ≥ 6  .

The smallest number that is greater than or equal to 6 is  6 .

The smallest piece that needs to be cut from each of the sticks is 6 inches long.

2.

1.  Choose values for  x  and then find  f(x)  and  g(x)  using that  x  value.

For the first  x  value I choose  0 . Then...

f(x)  =  f(0)   =   0² - 2(0) + 2  =  2

g(x)  =  g(0)  =  3(0) - 0²  =  0

For the second  x  value I choose  1 . Then...

f(x)  =  f(1)  =  1² - 2(1) + 2  =  1 - 2 + 2  =  1

g(x)  =  g(1)  =  3(1) - 1²  =  3 - 1  =  2

For the third  x  value I choose  1/2 . Then...

f(x)  =  f(1/2)  =  (1/2)² - 2(1/2) + 2  =  1/4 - 1 + 2  =  1/4 + 1  =  5/4

g(x)  =  g(1/2)  =  3(1/2) - (1/2)²  =  3/2 - 1/4  =  6/4 - 1/4  =  5/4

For the fourth  x  value I choose  2 . Then...

f(x)  =  f(2)  =  2² - 2(2) + 2  =  4 - 4 + 2  =  2

g(x)  =  g(2)  =  3(2) - 2²  =  6 - 4  =  2

Do you see how to do this? I'll let you pick the fifth  x  value.

Notice that when  x = 1/2 ,  f(x) = 5/4  and  g(x) = 5/4

So when  x = 1/2 ,  f(x) = g(x)

And when  x = 2 ,  f(x) = 2  and  g(x) = 2

So when  x  = 2 ,  f(x) = g(x)

The  x  values that make  f(x) = g(x)  are  1/2  and  2 .

So the  x  values that are solutions to the system are  1/2  and  2 .

area of sector   =   (1/2) * (radius)² * measure of central angle in radians

area of sector   =   (1/2) * 12² * 2.3      sq. meters

area of sector   =   (1/2) * 144 * 2.3      sq. meters

area of sector   =   72 * 2.3     sq meters

area of sector   =   165.6        sq meters

I think I see what you did......remember  12² = 12 * 12   ....NOT....  12 * 2     

A)   ∠A  intercepts arc CD.

B)   ∠BED  intercepts arc BCD.

C)   The measure of an inscribed angle is half the measure of its intercepted arc. So...

       The measure of ∠BED is half the measure of arc BCD

       The measure of ∠BED   =   (1/2)(the measure of arc BCD)

       85°   =   (1/2)(the measure of arc BCD)

       2(85°)   =   the measure of arc BCD

       170°   =   the measure of arc BCD

D)   The measure of ∠A   =   (1/2)(the measure of arc CD)

       The measure of ∠A   =   (1/2)(128°)

       The measure of ∠A   =   64°

E)   The sum of all the arcs in a circle is 360° , so...

       The measure of the rest of the circle   =   360° - 82°

       The measure of the rest of the circle   =   278°

The aspect ratio of a computer monitor is given as width : height.   So....

width : height  =  16 :9

We are given that the width is 20 inches

Let the height be  h  inches.

20 inches : h inches  =  16 : 9

20 : h  =  16 : 9

                              This means the same as..

20 / h  =  16 / 9

                              Multiply both sides of the equation by  h .

20  =  h * 16/9

                              Multiply both sides of the equation by  9/16 .

20 * 9/16  =  h

11.25  =  h

Now we can use the Pythagorean Theorem to find the length of the diagonal.

(width)² + (height)²  =  (length of diagonal)²

20² + 11.25²  =  (length of diagonal)²

400 + 126.5625  =  (length of diagonal)²

526.5625  =  (length of diagonal)²

√526.5625  =  length of diagonal

length of diagonal   ≈   23  inches

The radius of circle 1 is  8x

The radius of circle 2 is  2x

The radius of circle 1 is  4  times the radius of circle 2, so

Circle 1 is a dilation of circle 2 with a scale factor of 4.