geno3141

(-7i)·(8i)  =  (-7·8)·(i·i)  =  (-56)·(i²)  =  (-56)·(-1)  =  56  =  56 + 0i

If h(x)  =  -3x² + 1,

then  h'(x)  =  -6x

Thus,  h'(-2)  =  -6(-2)  =  12  <--  slope when x = -2.

(2/3)(12 - 21c)

=  (2/3)(12) - (2/3)(21c)

=  8 - 14c

outs?   oats?

Total number of grams:  250g + 150g + 100g  =  500g

a)  250g / 500g  =  1/2

b)  150g / 500g  =  3/10

Two of the points on the graph are (2, -2)  and (3, -8)

f(x)  =  a^x/b

(2, -2)   --->   a²/b  =  -2     --->   a²  =  -2b     --->    a²/-2  =  b

(3, -8)   --->   a³/b  =  -8     --->   a³  =  -8b     --->    a³/-8  =  b

Setting the first equation equal to the second:     a²/-2   =   a³/-8 

Multiplying both sides by -8:                               4a²   =   a³

Dividing both sides by  a²  (This step is allowed provided that a²  ≠ 0):       4  =  a

Since  a²/b  =  -2  and  a  =  4:  16/b  =  -2     --->   b  =  -8

Now, find a + b:

As  x→∞,  51^x→∞  and  2·51^x→∞. Subtracting this from 3 gives values that approach -∞.

As  x→-∞,  51^x→0  and  2·51^x→0.  Subtracting this from 3 gives values that approach (but does not quite reach) 3.

The range is  (-∞, 3)

This problem would give a "nice" answer if 64 were a power of 3; however, ...

Problem:         81^x  =  64

Take the log of both sides:                  log(81^x)  =  log(64)

Exponents come out as multipliers:    x·log(81)  =  log(64)

--->                                                           x  =  log(64) / log(81)  ≈  610.940

27^{x + 1}  =  27^{610.940 + 1}  =  27^611.940

interest = principal x rate x time

End of the first year:

--->     interest earned  =  $10,000 x 0.10 x 1  =  $1000

--->     total value  =  $10,000 + $1,000  =  $11,000

End of the second year:

--->    interest earned  =  $11,000 x 0.10 x 1  =  $1100

--->     total value  =  $11,000 + $1100  =  $12,100.

End of the third year:

--->     interest earned  =  $12,100 x 0.10 x 1  =  $dddd

--->     total value  =  $12,100 + $dddd  =   $answer

Write  3^{2c + 1}  as  (3^2c) · (3)  or as 3·3^2c

This problem becomes:  3·3^2c  =  28·3^c - 9

which is:                      3·3^2c - 28·3^c + 9  =  0

I'm going to replace  3^c  with  x  and  3^2c  with x²:

--->                            3x² - 28x + 9  =  0

--->     Factor:             (3x - 1)(x - 9)  =  0

Which means:              3x - 1  =  0         or         x - 9  =  0

--->                                   x  =  1/3      or              x  =  9

But   x =  3^c  --->              3^c   = 1/3      or             3^c  = 9

--->                                 3^c   =  3^-1      or             3^c  = 3²

So:                                   c  =  -1        or               c  =  2

If you have a unit circle, the value of sine is the y-value, the value of cosine is the x-value, and the value of tangent is y/x.

So, if the point is  x = 3/5  and  y = 4/5, the value of tan θ  =  (4/5) / (3/5)  =  4/3