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f(x)  =  [ lnx + log₂ x ] / [tanx]    =

lnx / tanx    +  log₂ x / tanx      using the Quotient Rule  on both, we have

[ (1/x) * tanx -  lnx * sec^2 x]  / [tan^2 x]     + [ 1 / [x ln 2] * tanx - log₂ x * sec^2x] / [tan^2x]  =

[ tanx/x - lnx * [tan^2 x + 1] / [tan^2 x]  + ( [ tanx/ [ x ln2] -  log₂ x [ 1 + tan^2x] ) / [tan^2 x]  =

[ tanx ( 1/x + 1/[ x ln2] )  - [ tan^2x + 1] ( lnx + log₂ x ) ] /  [ tan^2 x]  =

[ ln 2 + 1] / [ x* ln2 * tanx]  -  [ sec^2x / tan^2x] * (lnx + log₂ x)  =

[ ln 2 + 1] / [ x* ln2 * tanx]   -  csc^2x * (ln x + log₂ x )  =

[ln2 + 1]  * cot x  / [x ln 2]  -  csc^2x (ln x + log₂ x)

Good night. Sweet dreams hopefully LOL

y =  x^(2cosx)

lny = lnx^(2cosx)

lny  = 2cosx* [ln x]       [ use the product rule to differentiate the right side ]

y' / y  =  -2sinx*lnx - 2cosx[ 1/x]

y' / y  =  [-2*x*sinx*lnx - 2cosx] / x         multiply both sides by y  [ and y = x^(2cosx)]

y ' =  x^(2cosx) * [ -2*x*sinx*lnx - 2cosx] * [x^(-1)]

y '  =  -2 * [ x^(2cosx)* x^(-1) ] [ x*lnx*sinx - cosx] 

y '  = -2 * [ x^(2cosx - 1) ] * [x*nx*sinx - cosx]

You may be in my class because I have the same question on my hw

I assume this is supposed to be:

6x^2 - 13x + 5 = 0

(3x - 5) (2x - 1)   =  0     

Setting each factor to 0 and solving

3x - 5  = 0                           2x - 1     = 0

add 5 to both sides            add 1 to both sides

3x  = 5                                2x   =  1

divide both sides by 3        divide both sides by 2

x  = 5/3                    or         x  = 1/2

7and 3 fifths

We can find the horizontal distance, h,  from the plane to the ship as follows :

tan (40) =  1300 / h      .....rearrange as.....

h  = 1300/tan(40)   = about 1549.3 m

We need to find how many terms we have, as follows

1701  = 7(3)^n-1     divide both sides by 7

243 = 3^n-1   and 3⁵  = 243   so n = 6

So we have

S_n = a₁ [ (1-rⁿ) / (1 - r) ]

S₆ = 7 [ (1 - 3⁶) / (1 - 3) ]  =  7 [ 1 - 729] [ 1 - 3]  =  7 [-728] / [-2]  = 2548

z = the fraction 1 divided by 14. Or 1 over 14.

That would be correct.

The answer is 9:16, beacuse 3*3=9, 4*4=16.

The answer is 96

25=286.93(9.738^x)   divide both sides by 286.93

25/286.93  = 9.738^x     take the log of each side

log [ 25/286.93]  = log [9.738]^x    and we can write

log [ 25/286.93]  = x * log [9.738]     divide both sides by log [9.738]

log [ 25/286.93]  / log [9.738]  = x   =  about   -1.072

We need to know how many terms are to be summed......obviously, this sequence would have an unbounded infinite sum because each successive term is three times the previous one....

We need to solve this, first

192 = 3(2)^n-1     divide both sides by 3

64  = 2^n-1         and 2⁶ = 64,   so n = 7

So we  have the sum  of 7 terms

S_n =  3 [ (1 - 2⁷) / (1 - 2) ]    =   3 [ (1 - 128) / (-1) ] =    3* 127 =  381

This was answered here..... http://web2.0calc.com/questions/two-geometric-means-are-inserted-between-6-and-384-so-the-four-numbers-form-a-geometric-sequence-what-are-these-two-numbers

384/6 = r³

64  = r³       take the cube root of each side

4 = r

So the second number  = 6(4)  = 24

And the third = 24(4)  = 96

an = a1(r)n − 1=384=6(r)^4-1=4 common ratio.

The common is 4 and so we have:

6, 24, 96, 384..................................

The answer is C.

2x+3y=9    (1)

x+5y=8     [ multiply this eq. by -2]  =   -2x - 10y  = -16  .....  add this to the first eq.

0 - 7y  = -7

-7y  = -7      divide both sides by  -7    and y  = 1

Using (1) to find x

2x + 3(1) = 9

2x + 3  = 9     subtract 3 from both sides

2x = 6      divide both sides by 2

x = 3

So  (x , y)   = (3, 1)

you can't divide ANYTHING by zero...it is 'undefined'

Add the two equations together to 'eliminate'   'y'    then solve for 'x'

5x-y +3x + y = 0 + 24

8x =24

x=3                             Then substitute this value of 'x' into one of the equations to solve for 'y'.....

12    =  a₁(r)²   →    a₁  = 12/r²      (1)
-324 =  a₁(r)⁵     (2)

Putting  (1)  into (2), we have

-324  = (12/r²)(r)⁵

-324/12 = r³

-27  = r³    

r = (-27)^1/3 = -3

So

a₁  = 12/(-3)²  =  12/9  =  4/3