geno3141

If the problem is:  log₄ [ (x + 1)³ / 2 ]  =  22

Change from log form into exponential form:  (x + 1)³ / 2  =  4²²

Multiply both sides by 2:                                      (x + 1)³   =  2 · 4²²

Change the form of 4²²                                       (x + 1)³   =  2 · ( 2 · 2 )²²

                                                                            (x + 1)³   =  2 · ( 2² )²²

                                                                            (x + 1)³   =  2 · 2⁴⁴

Simplify:                                                               (x + 1)³   =  2⁴⁵ 

Find the cube root of both sides:                             x + 1  =  2¹⁵

Subtract 1 from both sides:                                           x  =  2¹⁵ - 1  =  32768 - 1  =  32767

The cube root of 2⁴⁵ is  (2⁴⁵)^1/3 =  2^{45 x 1/3}  =  2¹⁵.

The slash through the 'greater than or equal' sign means 'not (greater than or equal)' or, simplly, 'less than'.

Therefore, your inequality reduces to:  -x - 5  <  -1

Add 5 to both sides:                                    -x < 4

Divide both sides by -1:                                x > -4

(When you multiply or divide both sides of an inequality by a negative number, change the direction of the inequality sign.)

Since the domain is the set of negative integers, the answers include -3, -2, and -1.

Place a filled-in circle on -3, on -2, and on -1.

Each right triangle has 2 legs, the 2 sides that aren't the hypotenuse (the hypotenuse is the side opposite the right angle).

Starting with                                   x² / 2  =  7x

Multiplying both sides by 2:                 x²  =  14x

Subtract 14x from both sides:     x² - 14x  =  0     (which is answer 'c')

Answer 'a' is also algebraicly correct but, because the leading coefficient is negative, 'c' is preferred.

I don't believe that you have enough information because you can create more that one function.

y  =  -x(x + 4)²(x - 4)         y  =  x²(x + 4)(x - 4)       etc.

These functions can also have other non-zero coefficients:  y  =  -3x(x + 4)²(x - 4)            y  =  5x²(x + 4)(x - 4)

sin(x) + cox(x)  =  3/2

Square both sides:  [ sin(x) + cos(x) ]²  =  [ 3/2 ]²

Multiply out:               sin²(x) + 2sin(x)cos(x) + cos²(x)  =  9/4

Rearranging:             sin²(x) + cos²(x) + 2sin(x)cos(x)  =  9/4

Since  sin²(x) + cos²(x)  =  1:           1 + 2sin(x)cos(x)  =  9/4

Subtracting 1 from both sides:               2sin(x)cos(x)  =  5/4

Since  sin(2x)  =  2sin(x)cos(x):                       sin(2x)  =  5/4 

But the sine value can never be greater than 1   --->     sin(2x)  can't be 5/4     --->     There is no answer.

                     --->     This is an "impossible" problem.     <---

Your answers to #6 are correct.

8)  U  =  {1, 2, 3, 4, 5, 6}     A  =  {2, 3, 6}

c)  A'  =  {1, 4, 5}     --->     A ^A'  =  {2, 3, 6} ^ {1, 4, 5}  =  { }      (the null set)

d)  U'  =  { }     --->     U' ^ A  =  { } ^ {2, 3, 6}  =  { }

9)  P  =  {1, 3, 5, 7}     Q  =  {3, 7, 9, 11}     R  =  {1, 5, 8, 11}

a)   Show that:  P v Q  =  Q V P

      --->    Since  P v Q  =  {1, 3, 5, 7} v {3, 7, 9, 11}  =  {1, 3, 4, 7, 9, 11}

      --->     and   Q v P  =  {3, 7, 9, 11} v {1, 3, 5,7 }  =  {1, 3, 4, 7, 9, 11},

                therefore  P v Q  =  Q v P

b)  Show that:  (P v Q) v R  =  P v (Q v R)

     --->  Since (P v Q) v R  =  ( {1, 3, 5, 7} v {3, 7, 9, 11} ) v {1, 5, 8, 11}  

                                         =  {1, 3, 5, 7, 9, 11} v {1, 5, 8, 11}  =  {1, 3, 5, 7, 8, 9, 11}

     --->  and  P v (Q v R)  =  {1, 3, 5, 7} v ( {3, 7, 9, 11} v {1, 5, 8, 11} )

                                        =  {1, 3, 5, 7} v {1, 3, 5, 7, 8, 9, 11}  =  {1, 3, 5, 7, 8, 9, 11}

             therefore (P v Q) v R  =  P v (Q v R)

c)  Show that:  P ^ Q  =  Q ^ P

     --->  Since P ^ Q  =  {1, 3, 5, 7} ^ {3, 7, 9, 11}  =  {3, 7}

     --->   and  Q ^ P  =  {3, 7, 9, 11| ^ {1, 3, 5, 7}  =  {3, 7}

             therefore  P ^ Q  =  Q ^ P

d)  Show that:  (P ^ Q) ^ R  =  P ^ (Q ^ R)

     --->  Since  (P ^ Q) ^ R  =  ( {1, 3, 5, 7} ^ {3, 7, 9, 11} ) ^ {1, 5, 8, 11}

                                           =  {3, 7} ^ {1, 5, 8, 11}  =  { }

             and  P ^ (Q ^ R)  =  {1, 3, 5, 7} ^ ( {3, 7, 9, 11} ^ {1, 5, 8, 11} )

                                        =  {1, 3, 5, 7} ^ { }  =  { }

             therefore  (P ^ Q) ^ R  =  P ^ (Q ^ R)

e)  Show that  P v (Q ^ R)  =  (P v Q) ^ (P v R)

     --->  Since  P v (Q ^ R)  =  {1, 3, 5, 7} v ( {3, 7, 9, 11} ^ {1, 5, 8, 11} )

                                          =  {1, 3, 5, 7} v { 11  }  =  {1, 3, 5, 7, 11}

             and  (P v Q) ^ (P v R)  =  ( {1, 3, 5, 7} v {3, 7, 9, 11} ) ^ ( {  1, 3, 5, 7} v {1, 5, 8, 11} )

                                                 =  {1, 3, 5, 7, 9, 11} ^ {1, 3, 5, 7, 8, 11}  =  {1, 3, 5, 7, 11}

             therefore  P v (Q ^ R)  =  (P v Q) ^ (P v R)

f)  Show that  P ^ (Q v R)  =  (P ^ Q) v (P ^ R)

     --->  Since  P ^ (Q v R)  =  {1, 3, 5, 7} ^ ( {3, 7. 9. 11} v {1, 5, 8, 11} )

                                           =  {1, 3, 5, 7} ^ {1, 3, 5, 7, 8, 9, 11}  =  {1, 3, 5, 7 }

             and  (P ^ Q) v (P ^ R)  =  ( {1, 3, 5, 7} ^ (3, 7, 9, 11} ) v ( {1, 3, 5, 7} ^ {1, 5, 8, 11} )

                                                 =  {3, 7} v {1, 5}  =  {1, 3, 5, 7}

If the prism is a rectangular prism (a box), then the volume is found by multiplying the length times the width times the height.

V  =  25 x 50 x 2  =  2500 cubic meters.

[ sin(26) x 5 ] / [ sin(121) ]  =  2.557...