Question

f(x) = sin^4(x+pi/6) + cos^4 (x+pi/6) is

a) not a periodic fn
b) a periodic fn with period 2pi
c) a periodic fn with period pi
d) a periodic fn with period pi/2

2. For the hyperbola
9x^2-16y^2-18x+32y-151=0

a) one of the directrix is x =21/5
b) length of latus rectum = 9/2
c) focii are (6,1) & (-4,1)
d) ecentricity is 5/4

Answer 1

help........

Answer 2

any1 help me........

Answer 3

ishaan cann u

Answer 4

sebentar yahhhh tak solve dulu sabar ya jang

Answer 5

noras please type in english

Answer 6

waduh soryy je aku ndak iso boso londho je..iki boso jowo lho..indonesia made in punya lah

Answer 7

please type in english noras

Answer 8

hmmmmmmm

Answer 9

1) \[f(x)=\sin^4(x+\pi/6)+\cos^4(x+\pi/6)=\]\[=\sin^4(x+\pi/6)+2\sin^2(x+\pi/6)\cos^2(x+\pi/6)+\cos^4(x+\pi/6)-\]\[-2\sin^2(x+\pi/6)\cos^2(x+\pi/6)=(\sin^2(x+\pi/6)+\cos^2(x+\pi/6))^2-\]\[-2\sin^2(x+\pi/6)\cos^2(x+\pi/6)=1-\frac{1}{2}\sin^2(2x+\pi/3)=\]
\[=1-\frac{1}{2}\frac{1-\cos{(4x+2\pi/3)}}{2}=\frac{3}{4}+\frac{1}{4} \cos{(4x+2\pi/3)}\]
period is \[T=2\pi/4=\pi/2\]

Answer 10

2) \[9x^2-16y^2-18x+32y-151=0\]
\[9x^2-18x+9-16y^2+32y-16-144=0\]
\[9(x-1)^2-16(y-1)^2=144\]
\[\frac{(x-1)^2}{4^2}-\frac{(y-1)^2}{3^2}=1\]

Answer 11

u on facebook nikvist

Answer 12

I haven't got account on facebook

Answer 13

gmail ?

Answer 14

yes, nikvist83@gmail.com

Answer 15

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