help ! add maths :(

[Diishiik Douw]

2^ (y+1) = 0.2
solve

 

[HarisShahzad]

2^ (y+1) = 0.2
take log of both sides

log 2^(y+1)=log 0.2

(y+1) log 2=log 0.2

y+1=log 0.2/log 2
y+1= -2.3219
y= -2.3219 -1
y= -3.3219
y= -3.32 (3 significant figures)

 

[devilraj2505]

2^(y+1)=0.2
log0.2/log2=y+1
y=(log0.2/log2)-1

 

[Diishiik Douw]

Given that x is measured in radians and x > 10, find the smallest value of x such that:

10 cos ( x+ 1 /2 ) =3

 

[Diishiik Douw]

( 1 + sec x)(cosec x – cot x) = tan x

 

[HarisShahzad]

10 cos ( x+ 1 /2 ) =3

x > 10
x+1/2 > 10+1/2
x+1/2 > 21/2 or 10.5

cos ( x+ 1/2 ) = 3/10

x+1/2 = cos^-1 (3/10)
since cos is positive it is in first and third quadrant

x+1/2 = 1.266, 5.017, 7.546

since 1.266 is smallest hence

x+1/2 = 1.266
x= 1.266-1/2
x= 0.766

 

[HarisShahzad]

( 1 + sec x)(cosec x – cot x)
open brackets

cosec x -cotx + sec x cosec x -secx cotx

convert all to sin and cos

1/sinx - cos x/sinx + 1/sinx cosx - cosx/sinx cosx

in last expression - cosx/sinx conx. cosx cancels leaving -1/sinx which cancels with the first expression +1/sinx

leaving

-cos x /sinx + 1/sinx cosx

take LCM

-cos^2 x +1/cosx sinx

1-cos^2 x = sin^2 x
so

sin^2x / cos sinx

sinx cancels with sin^2 x and leaving

sinx / cosx =tanx