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2⁵ = 32

log₂ 32 = 5

"B"

thank you so much!

6e^4x - 2 = 3    add 2 to both sides

6e^4x  = 5        divide both sides by 6

e^4x  =  5/6      take the ln  of both sides

ln e^4x  = ln ( 5/6 )

(4x) ln e = ln (5/6)    and lne = 1 ....  so we have ......

4x  = ln (5/6)     divide both sides by 4

x  =   [ ln (5/6) ] / 4  = -0.0455803891984886 = -0.046

So "D" is correct

40+25x=15x+30  ........subtract 30 and 25x from both sides

40 - 30  =  15x - 25x

10  =  -10x        divide both sides by  -10

-1 = x

Tried it out but not getting the answer from the book. Answer to the problem for the three unknowns is 326, 34, 53. All ma measurements. 0.326, 0.034, 0.053 Amps converted from the ma's. That's the results on page 61 of 'Electronics A Systems Approach'

Probably error on my part, doing something wrong! But the figures i'm getting are nowhere near the answers.

"A" is correct

Here's a link to the graph:

https://www.desmos.com/calculator/nqjyarhcx8

 Hey.....Nataszaa...this is the first time I've tried to link a graph...let me know if you can access it!!!!

"A" is correct, Nataszaa

3ln3 + 3lnc  =

ln 3 ³ + ln c ³ =

ln 27 +  ln c ³ =

ln ( 27 c ³ )

"D"

sweet! i appreciate it CPhill!

"B" is correct, Nataszaa

((7w-6)/2)=((8w+3/3)+1    multiply both sides by the common denominator of the two fractions (6)

3[7w - 6 ] = 2[ 8w + 3]  + 6    simplify

21w - 18 = 16w + 6 + 6

21w - 18  = 16w + 12      add 18 to both sides and subtract 16w from both sides

5w = 30      divide both sides by 5

w = [30 / 5]  = 6

4 ^-2 = 1/16

D is correct

3 ln x - 2 ln c =

ln x ³ -  ln c ² =

ln ( x ³/ c ² )

"a" is correct......

Filling what we know, we have

6 = t / (3.3log₈ 7681)    ....and, using the "change of base" rule, we can write

6 = t / [(3.3)log 7681/log8)     .......and multiplying both sides by the denominator on the right, we have

6(3.3)log7681/log8 = t ≈ 85 hrs (rounded)

thanks so much CPhill!

There you go, Nataszaa

Thanks for the tips Alan. For the 3 x 1 vector i've never worked with those yet or studied anything about them so i didn't know i could use that option for doing the final multiplication of the inverse and constant matrices. I thought i always had to have matrix selected for both sides!!

I'll look into the SMath tip and look it up via google as it sounds like what i'm looking for.

Thanks again for all your help.

I'm making a guess as to what is meant here

x [√(9-4)/3] = 5   multiply both sides by 3

x √5 = 15       divide both sides by √5

x = 15/√5 = (15√5) / 5   =  3√5

sin 240 x sec (-45)-tan(-30) x csc(-315)=

[-√(3)/2]*[√2] - [-1/√(3)]*[√2] =

[-√6/2] + [√2/√3]  =

[-√6/2] + [√6/3]  =

[-3√6]/6 + [2√6]/6  =

-√6 / 6

You should be able to do a 3x1 vector - it's the default when you switch into the vector/matrix mode!

I think there's a bug in the calculator that prevents the A^-1*B option working directly.  I've sent a message to Admin about it.   ADMIN HAS NOW FIXED THIS!

I wouldn't use Cramer's rule for anything more than a 3x3.  I wouldn't expect anyone to do a 4x4 or bigger by hand!  Get hold of some software that will do this for you.  There is a free download program called SMath Studio Desktop that should do this. 

hi Melody,

new feature: "post-date" on the right side is linked to the address of the post

yes!its some complicated math!ive seen many like these!

too many parenthisis and irrevelent equations.......or maybe its some complicated math i dont know?

That was really a nice answer sman3 ! you know i had almost forgotten how to convert percentage into fraction , i had learnt this very long ago!

thanks for reminding me about it!

thumbs up for you!

Your Welcome Queen!

And special Welcome backs are only for special people!but i think still some special people arent back!seems they r all very busy but yeah let them enjpy their last holidays!u know in summer out here when we get our summer holidays its so hot that we cant even enjoy them by going somewhere!

But my dad's planning to take us somewhere in the coming holidays!out here winter vacations r much fun!i love winters!