geno3141

1)  Find the midpoint of the two points.

2)  Find the slope of the line that connects those two points.

3)  Find the negative reciprocal of the slope of step 2). This will be the slope of the perpendicular bisector.

4)  Use the point-slope form for the equation of the line (the perpendicular bisector).

     Use the answer to step 1) for the point.

     Use the answer to step 2) for the slope.

5)  Rewrite the equation of step 4) into y = mx + b form.

6)  From step 5), pick out the value of m and the value of b.

7)  Add the two values of step 6) together to get the answer.

Let  L  represent the length of the rectangle and  W  represent the width.

L · W  =  32          2L + 2W  =  24   --->   L + W  =  12   --->   W  =  12 - L

L · W  =  32     --->     L · (12 - L)  =  32     --->     12L - L²  =  32

                       --->     0  =  L² - 12L + 32     --->     0  =  (L - 8)(L - 4)     

Length  =  8     and     Width  =  4.

The distance from the center of the circle to the top side = 2.

The distance from the center of the circle to the right side = 4.

The distance from the center of the circle to the top-right corner  =  r

Using the Pythagorean Theorem:  r²  =  2² + 4²  =  4 + 16  =  20    --->     r  =  sqrt(20)

The area of the circle is:  pi · sqrt(20)²  =  20pi

The area of the rectangle is  8 · 4  =  32

The area of the shaded region  =  20pi - 32

Let  A  =  center of the top-left circle.

Let  B  =  center of the top-right circle.

Let  C  =  center of the bottom circle.

Draw AB; label the point of tangency of the two circle  X.

Draw AC; its length is 24 + 16  =  40.

AX  =  24.

Use the Pythagorean Theorem to find XC:  XC² + XA²  =  AB²     --->     XC² + 24²  =  40²

Solving:  XC  =  32.

The width of the rectangle is  24 + 24 + 24 + 24  =  96

The height of the rectangle is:  24 + 32 + 16  =  72

From these two dimensions, you can find the area.

The fourth shoe will guarantee that there is a matching pair.

What is the number?

If  (a + b) / b  =  5     --->     a/b + b/b  =  5     --->     a/b + 1  =  5     --->     a/b  =  4

If  (b + c) / c  =  4     --->     b/c + c/c  =  4     --->      b/c + 1  =  4      --->     b/c  =  3     --->     c/b  =  1/3

a/b + c/b  =  4 + 1/3  =  13/3     and     a/b + c/b  =  (a + c)/b     --->     (a + c)/b  =  13/3

Since  b/c  =  3     --->     b  =  3c

(a + c)/b  =  13/3     and    b  =  3c     --->     (a + c)/(3c)  =  13/3     --->     (a + c)/c  =  13  

x³ - x² - 21x + 45  can be factored into  (x + 5)(x -3)²

so multiply each term by   (x + 5)(x -3)² 

You'll end with:      1  =  A(x - 3)² + B(x - 3)(x + 5) + C(x + 5)

                              1  =  A(x² - 6x + 9) + B(x² + 2x - 15) + C(x - 5)

                              1  =  Ax² - 6Ax + 9A + Bx² + 2Bx - 15B + Cx - 5C

             0x² + 0x + 1  =  Ax² + Bx²  - 6Ax + 2Bx + Cx + 9A - 15B - 5C

So:            Ax² + Bx²  =  0x²   --->   (A + B)x²  =  0x²   --->    A + B  =  0

       -6Ax + 2Bx + Cx  =  0x     --->   (-6A + 2B + C)x  =  0x   --->   -6A + 2B + C  =  0

           9A - 15B - 5C  =  1   

Now, solve these three equations simultaneously:         A +     B          =  0

                                                                                     -6A +   2B +  C  =  0

                                                                                       9A - 15B - 5C   =  1   

There are 300 spiders and millipedes, of which 10% are spiders:  0.10 · 300  =  30.

He has 30 spiders and 270 millipedes.

He wants a group of millipedes and spiders to contain 30 spiders and these 30 spiders 

are 20% of the group.

Let  x  be the total number of millipedes and spiders:  0.20 · x  =  30

Dividing both sides by 0.20:                                                    x  =  30 / 0.20  =  150.

He'll end with 30 spiders and 120 millipedes.

He'll need to remove  270 - 120  =  150  millipedes.

Where's the diagram?

Did you enter the problem correctly?