Question

pls help

Answer 1

the file is here

Answer 2

does hg(x) mean h(g(x))?

Answer 3

I).. g=1/x.......g=1/(1/2)=2
then h=x^2. 2^2=4
hg(1/2)=4

Answer 4

@Nnesha can u answer the other questions

Answer 5

@ganeshie8 @Abhisar plss help me

Answer 6

what do u mean

Answer 7

i dont know

Answer 8

@butterflydreamer do u know the answers pls help me

Answer 9

@dan815

Answer 10

@wio @iambatman help pls

Answer 11

@Hero @Callisto @Jhannybean @Miracrown help me i will fan and give a medal

Answer 12

@shrutipande9 @Preetha @nincompoop @MARC_

Answer 13

\[hg (\frac{ 1 }{ 2 })=h(g(\frac{ 1 }{ 2 }))\]\[=h(1\div \frac{ 1 }{ 2 })\]\[=(2)^2\]\[=?\] solve it @hari5719

Answer 14

4

Answer 15

YUP

Answer 16

its the rest that i dont understand the a2 and the rest @MARC_

Answer 17

\[fhg(\frac{ 1 }{ 2 })=fh(g(\frac{ 1 }{ 2 }))\]\[=fh(1\div \frac{ 1 }{ 2 })\]\[=fh(2)\]\[=f(h(2))\]\[=f((2)^2)\]\[=f(4)\]\[=3(4-1)\]\[=3\times3\]\[=?\] solve it @hari5719

Answer 18

9

Answer 19

yup

Answer 20

how about inverse

Answer 21

\[f(x)=3(x-1)\]\[f(y)=3(x-1)\]\[y=3(x-1)\]\[3y=x-1\]\[3y+1=x\]

Answer 22

Therefore,\[f^-1:x \rightarrow3y+1\]

Answer 23

\[hfg(x)=36\]\[hf(g(x))=36\]\[hf(\frac{ 1 }{ x })=36\]\[h(\frac{ 1 }{ 3(x-1) })=36\]\[(\frac{ 1 }{ 3(x-1) })^2=36\]\[\frac{ 1 }{ 3(x-1) }=\sqrt{36}\]\[\frac{ 1 }{ 3(x-1) }=6\]\[\frac{ 1 }{ 3x-3 }=6\]\[1=6(3x-3)\]\[1=18x-18\]\[1+18=18x\]\[19=18x\]\[x=?\] solve it @hari5719

Answer 24

1.05
so this is how i was breaking my head on it how do i do the equeation

Answer 25

make it in fraction @hari5719

Answer 26

19/18

Answer 27

yup :)

Answer 28

what about the equation one @MARC_

Answer 29

which one ? @hari5719

Answer 30

last

Answer 31

the question ask to solve x

Answer 32

\[x=\frac{ 19 }{ 18 }\]

Answer 33

thats it well thanx @MARC_

Answer 34

np