geno3141

sun(x) + sin(x)·tan²(x) 

=  sin(x)[ 1 + tan²(x) ] 

=  sin(x)[ sec²(x) ] 

=  sin(x)·sec(x)·sec(x)

=  sin(x)·[ 1/cos(x) ]·sec(x)

=  [ sin(x)/cos(x) ]·sec(x)

=  tan(x)·sec(x)

Find the equation  f(x)  =  a^x + b

For  (-2,-2)  --->   -2  =  a^-2 + b   --->   -2 - a^-2  =  b

For  (-1,-4)  --->   -4  =  a^-1 + b   --->   -4 - a^-1  =  b

Setting these two equation equal to each other:  -2 - a^-2  =  -4 - a^-1 

                                                                              2 + a^-2  =  4 + a^-1

Multiplying by a²:                                                 2a² + 1  =  4a² + a

                                                                                      0  =  2a² + a - 1

                                                                                      0  =  (2a - 1)(a + 1)

So, either  a = ½  or  a = -1.

If  a = ½   --->   -4  =  a^-1 + b   --->   -4  =  (½)^-1  + b   --->   -4  =  2 + b   --->   b  =  -6   

   --->   f(x)  =  ½^x - 6

If  a = -1   --->   -4  =  (-1)^-1 + b   --->   -4  =  -1  +  b   --->   b  =  -3

   --->   f(x)  =  (-1)^x - 3                [Note that "-1" must be in parentheses; this may eliminate it from being

                                                          a posible answer.]

If he has 1 1/2 pounds and another 3 1/10 pounds, he has (1 1/2 + 3 1/10) pounds  =  (4 + 1/2 + 1/10) pounds

  1/2 + 1/10  (using the common denominator of 10)  =  5/10 + 1/10  =  6/10

So, he has  4 3/5 pounds; this is more than he needs.

Divide the coefficient of the x-term by 2 and square the answer:

     (-5/2)²  =  25/4

1)  sin(-x)  =  -sin(x)  

     tan(-x)  =  -tan(x)

     sin(-x) · tan(-x)  =  -sin(x) · -tan(x)  =  sin(x) · tan(x)

The volume of a box is  length x width x height.

The original box has a volume of 10.5 x 4.5 x 8  =  378  cu in.

Find the volume of the boxes for each posibility byusing this formula and adding the answers. 

If the answer is at least as large as 378 cu in, that answer is a possibility.

Since two of the balls have the number 4, I'm going to call them 4₁ and 4₂.

This means, that there are these four balls:  2, 4₁, 4₂, and 5.

Possibilities:

   2 + 4₁  =  6        4₁ + 2   =  6          4₂ + 2  =  6          5 + 2    =  7

   2 + 4₂  =  6        4₁ + 4₂  =  8         4₂ + 4₁  =  8         5 + 4₁  =  9

   2 + 5    =  7       4₁ + 5    =  9         4₂ + 5    =  9         5 + 4₂  =  9

There are 4 ways to get a total of 6.

There are 2 ways to get a total of 7.

There are 2 ways to get a total of 8.

There are 4 ways to get a total of 9.

Can you finish the problem?

Solve for s:                                      t²  =  2p + as 

Rewrtie:                                 2p + as  =  t² 

subtract as from both sides:          2p  =  t² - as

divide by 2:                                      p  =  ( t² - as ) / 2

The answer to the first question is  A  =  pi·r²; but, in the future, it would be a lot quicker and easier for you to check your textbook or "google" the question.

For the second queston: draw a vertical line dividing the picture into two rectangles: a 15 x 30 rectangle on the left and a 12 x 15 rectangle on the right. 

Find those two areas and add them together.

The equaton  y = cos(x)  has a period of  2pi  or  360°.

This equation goes "six times as fast" which means that its period is only one-sixth the period of the base function.

So, take the base period and divide by 6.