how do i slove this: h(x)=(4+3x)^5?

Question

and this one too thanks.
h(x)=(2+x^3)^6
---------- <<<< that line is liek a fraction
( 2-x^3)

Answer 1

You need to give some directions. What is the problem asking you to do?
Note: The person below is incorrect. (4+3x)^5 does NOT equal
4^5 + (3x)^5

Answer 2

h(x) = ((2 + x^3)^6)/(2 - x^3)
h(x) = ((x^3 + 2)^6)/(-x^3 + 2)
h(x) = ((x^3 + 2)^6)/(-(x^3 - 2))
h(x) = -((x^3 + 2)^6)/(x^3 - 2)

for a graph, go to http://www.calculator.com/calcs/GCalc.html

type in -((x^3 + 2)^6)/(x^3 - 2)

other than that, there is nothing to solve or simplify

x can be anything except for cbrt(2)
y >= 0

Answer 3

All you do is raise 4 and 3 to the 5th power like this...

h(x)=1024 + 243x
^ ^
(4^5) (3^5)

then substitute a number in for x if you need to solve it completly

and for the 2nd one:

h(x)=64+x^18
^ ^
(2^6) (muptly 3 and 6 for exponents)
---------------------------------
2-x^3

Answer 4

NOTE:
If you can't see what is written put your mouse
pointer on the .... to see what is written

expand it out
h(x)=(4+3x)^5
h(x)=(4+3x)(4+3x)(4+3x)(4+3x)(4+3x)
h(x)=(243x^5+1620x^4+4320x^3+5760X^2+3840x+1024)
break it down so you won't get lost like this, but make sure you multiply five (4+3x).

(4+3x)(4+3x) = 9x^2+24x+16
(9x^2+24x+16)(4+3x) = 27x^3+108x^2+144x+64
(27x^3+108x^2+144x+64)(4+3x) =
81x^4+432x^3+864x^2+768x+256
(81x^4+432x^3+864x^2+768x+256)(4+3x) =
243x^5+1620x^4+4320x^3+5760X^2+3840x+1024 = h(x)

Answer 5

1.5 2

Answer 6

Umm, what's that weird curly thing?

Answer 7

i don't know how to "slove" anything....