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-4c - 11  =  4c + 21

Add  4c  to both sides:

     -11  =  8c + 21

Subtract  21  from both sides:

     -32  =  8c

Divide both sides by 8:

       -4  =  c

If you have a function  f(x)  =  5, the limit will be 5.

Rewrite  8t²  ≤  3 - 10 t  as      8t² + 10t - 3 ≤ 0

Factor:                                (4t - 1)(2t + 3) ≤ 0

Solve the related equality:     (4t - 1)(2t + 3)  =  0 

     If:       4t - 1  =  0     --->     4t  =  1     --->      t  =  1/4  

     If:      2t + 3  =  0     --->     2t  =  -3     --->     t  =  -3/2 

These two points split the number line into three groups:     t ≤ - 3/2     -3/2 ≤ t < 1/4     t ≥ 1/4

Pick any number for t such that   t ≤ - 3/2;  replace it into  8t² + 10t - 3  

     and check to see if that answer ≤ 0:

     I'm going to try -10:        8(-10)² + 10(-10) - 3  =  697;  since this number isn't ≤ 0, these numbers                                                                                               don't work.

Pick any number for t such that  -3/2 ≤ t < 1/4 

     I'm going to try 0:     8(0)² + 10(0) - 3  =  -3; since this aswer is less than zero, these numbers work.

Pick any number for  t  such that     t ≥ 1/4:

     I'm goint to try 1:     8(1)² + 10(1) - 3  =  15; since this answer isn't ≤ 0, these numbers don't work.

Answer:    -3/2  ≤  t <  1/4 

I also need help on this.

Upon graphing, there is a zero at x = 6.239 (approximately).

If you use the discriminant of the quadratic formula:  b² - 4ac to determine real from complex solution:

If b² - 4ac > 0  then there will be only real.

   a = 1, b = b, and c = 15

   b² -4(1)(15) > 0

   b² > 60

   b² - 60 > 0   can be solved by factoring:

  (b - √60)(b +√60) > 0

Two cases:  

     First case: when both are positive:  

          b - √60 > 0  and  b + √60 > 0    --->     b > √60  and  b > -√60     --->   b > √60

     Second case: when both are negative:

           b - √60 < 0  and  b + √60 < 0    --->     b < √60  and  b < -√60     --->   b < -√60

These are the cases when it works; so the time that it doesn't work are:

          -√60 < b < √60

So now, list all the integers between those two radicals.

9. · 8.  +  6.  +  4.  +  75. · 0. 

If you place this angle on a graph, it will be in the third quadrant, 45° below the x-axis.

The sides of a  45° - 45° - 90°  triangle are  1 - 1 - √(2)

Labeling the three sides: the hypotenuse is '1', the horizontal (x-value) is '-1', and the vertical (y-value) is also '-1'.

The value of sin is the y-value divided by the hypotenuse:  sin(225°)  =  - 1 / √(2).

Rationalizing the denominator,  sin(225°)  =  - √(2 / 2).

Let us say that each side of the original block has length 'x'; then its volume is  x³.

If you make each side twice as long, each side will have length '2x'; then its volume is (2x)³  =  8x³.

So, the new cube will have 8 times the volume of the old cube.

If the old cube weighs 6 pounds, the new cube will weigh 6 x 8  =  48 pounds.

It equals no1 really cares work it out yourself 69 69 69 69 GET REKD M9

9 + 10 is 21 you stupid 

c³ + 4c > 5c²

Rearrange:  c³ - 5c²+ 4c > 0

Factor:   c(c² - 5c + 4) > 0          ----->         c(c - 4)(c - 1 ) > 0

Each section could either be positive or negative:

If all the factors are positive, the answer would be positive:  c > 0  and  c - 4 > 0  and  c - 1 > 0

           which is this:  c > 0  and  c > 4  and c > 1    which is true only when c > 4.

If two of the factors are positive, the answer will be negative, so we don't have to consider this.

If only one of the factors is positive, the answer will be negative; there are three possibilities:

          c > 0  and  c - 4 < 0  and  c - 1 < 0   --->   c > 0  and  c < 4  and  c < 1   --->   0 < c < 1

          c < 0  and c - 4 > 0  and  c - 1 < 0   --->   c < 0  and c > 4  and c < 1  (impossible)

          c < 0  and  c - 4 < 0  and  c - 1 > 0   --->  c < 0  and  c < 4  and  c > 1   (impossible)

If none of the factors is positive, the answer will be negative, so we don't have to consider this.

Answer:  Combining the first answer with the third answer:    {c > 4}  or  {0 < c < 1} 

Yes; a negative is less than a positive; that's why the answer includes the word "or"; the answer is either in the first section (less than the negative) or in the second section (more than the positive); it can't be in both.

I think the answer that geno gave is incorrect. Sorry

Good luck in the next world!

isnt' the negative less than a positive? So how can x be less than the -10.965 and greater than 3.465?

Using deMoivre's formula:

(1cis60)^1991  =  (1^1991)cis(60*1991)  =  cis(119460)  =  cis(300)  =  .5 + -√(3)/2i

(1cis(-60))^1991  =  (1^1991)cis(-60*1991)  =  cis(-119460)  =  cis(-300)  =  .5 + √(3)/2i

Adding:   .5 + -√(3)/2i  +  .5 + √(3)/2i  = 1 + 0i  =  1

Consider three cases:

1) z = -1: this is impossible because the denominator would become zero.

2) z > -1  (This makes the denominator positive; so when you multiply both sides by z + 1, you do not

                     need to change the direction of the inequality.)

     (3 - z) / (z + 1) ≥ 1

     Multiply both sides by z + 1:

      3 - z ≥ 1(z + 1)

      3 - z ≥ z + 1

      2 - z ≥ z

         2  ≥ 2z

         1 ≥ z     --->  z ≤ 1 and z > -1     --->     -1 < z ≤ 1

3)  z  < -1  (This makes the denominator negative; so when you multiply both sides by z + 1, you have

                     to change the direction of the inequality.)

     (3 - z) / (z + 1) ≥ 1

     Multiply both sides by z + 1:

      3 - z ≤ 1(z + 1)

      3 - z ≤  z + 1

      2 - z ≥  z

         2  ≤  2z

         1 ≤  z     --->  z ≥ 1  (This is imposible since the assumption of this section was z  < -1.)

Conclusion:  the answer is:  -1 < z ≤ 1      <==============               

If you have a negative exponent in the numerator, write it in the denominator with a positive exponent.

Example:   x^-7  =  1 / x^7

Example:  3x^-4 / 14  =  3 / (14x^4)

Example:  3x^-2y^5z^-4  =  3y^5 / (x^2z^4)

If you have a negative exponent in the denominator, write it in the numerator with a positive exponent.

Example:  1 / x^-8  =  x^8

Example:  3x^3 / 2y^-6  =  3x^3Y^6 / 2

Example:  4x^5y^-2Z^-5 / (3x^-3Y^6z^-3)  =  4x^5x^3z^3 / (3y^2y^6z^5)  =  4x^8 / (3y^8z^2)

x(x + 7.5) > 38

x² + 7.5x - 38 > 0

First, find the solutions of  x² + 7.5x - 38 = 0

I used the quadratic formula and got  x =  3.465  and  x  =  -10.965

These two numbers divide the x-axis into three groups.

So, there are three groups  (x <  -10.965)        (-10.965 < x < 3.465)       (x > 3.465)

Try any number from the first group; I'll choose -20; and put it back into the original inequality:

       x(x + 7.5) > 38     ===>     -20(-20 + 7.5)  =  250 and that's > 38 go this section works.

Try any number from the second group; I'll choose 0; and put it back into the original inequality:

       x(x + 7.5) > 38     ===>     0(0 + 7.5)  =  0 and that's not > 38 go this section doesn't works.

Try any number from the third group; I'll choose 10; and put it back into the original inequality:

       x(x + 7.5) > 38     ===>     10(10 + 7.5)  =  175 and that's > 38 go this section works.

Since the first group and the third group both work, the answer is:  x < -10.965  or  x > 3.465.

For the product of two numbers to be less than or equal to zero:

1) either one can be zero; thus,  s = -4  or  s = 3.

2) the first one is negative and the second one is positive:  

              x + 4 < 0  and  s - 3 > 0

 so:         x  < -4  and  x  > 3     (but this situation isn't possible)

3) the first one is positive and the second one is negative:

              x + 4 > 0  and  s - 3 < 0

 so:         x  > -4  and  x  < 3        (this situation is possible)

Thus, the answer (from part 1 and part 3):  -4 ≤ x ≤ 3

Part a:

Domain:  set of all possible real numbers:  -∞ < x < ∞

Range:    the set of all real numbers less than or equal to 68:  -∞ < x < 68

Part b:

Domain: 0 ≤ x ≤ 4

Range:   0 ≤ y ≤ 68

Part c:

The graph is a model of the time that the ball is in the air and the height of the ball.

x represents time; it can't be less than 0, nor can it be greater than the time that the ball returns to earth (the bouncing after that is not part of this equation).

y represents height; it can't be less than 0, for the ball is not kicked out of a hole nor is it kicked into a hole.

These restrictions weren't put into the equation.

I'm not sure which way you mean the problem:

-(-1)³ · (2-4)² - (-100)(4)  =  -(-1) · (-2)² - (-400)  =  1·4 + 400  =  4 + 400  =  404

-(-1)³ · (2-4)² - (-100)^4  =   -(-1) · (-2)² - (100 000 000)  = 1 · 4 - 100 000 000  =  -99 999 996

Or, did you mean something else?

Because cos has a period of 2π; therefore, ...   cos(4π)  =  cos(2π)  =  cos(0)  =  cos(-2π)  =  cos(-4π)  ...