Additional Math Help HERE ONLY...!

[astarmathsandphysics]

HOW do u get y
   3^2y +5(3^y-10)=0

sub p=3^y to get p²+5(p-10)=0 so p²+5p-50=(p+10)(p-5)=0 so p=-10 or 5 so y=(logp)/log3 =(log-10)/log3 impossible since log-10 does not exist or y=(log5)/log3

[shan2391]

Thanks astar
By the way i was reading the syllabus n did not find velcity n relative velocity in it

[sweetsh]

3(sinx-cosx)=2(sinx+cosx)

?

[sweetsh]

PLease answer

[shan2391]

3sinx-3cosx=2sinx+2cosx
sinx=5cosx
sinx/cosx=5
tanx=5
x=78.69

[sweetsh]

Thank youuuuu

[moraesikae]

3sinx-3cosx=2sinx+2cosx
sinx=5cosx
sinx/cosx=5
tanx=5
x=78.69

there is more to it,,,it could be either 78.89 or 256.7

[shan2391]

GUys can any 1 help me with how to find domain and range of functions

[moraesikae]

3sinx-3cosx=2sinx+2cosx
sinx=5cosx
sinx/cosx=5
tanx=5
x=78.69

there is more to it,,,it could be either 78.89 or 256.7

sorry i mean 78.69.

[moraesikae]

GUys can any 1 help me with how to find domain and range of functions

read previous pages.. it's all there

[Sweet_03]

Thaaaank You Very Much

[ramezamgad]

3sinx-3cosx=2sinx+2cosx
sinx=5cosx
sinx/cosx=5
tanx=5
x=78.69

there is more to it,,,it could be either 78.89 or 256.7

sorry i mean 78.69.

hey sweetsh
the method is correct but after obtaining the eq tanx=5 u should see the range of values of x written in the quest. it is generally 0<x<360
so u have to think when tanx is +ve and tan is +ve in the 1st and 3rd quadrats (ASTC rule)
so after getting x of the 1st quadrat u calculate the other value in the 3rd quadrat which is 180+x
finally x={78.7,258.7}

[shan2391]

hey can any1 giv me a breif about how to find range & domain of a function.
AM really struggling wid this

[ramezamgad]

for a linear function u must have either the domain or the range and use it to find the other
the domain is the values of x and they produce values of y called the range

eg  find the range for the eq. f(x)=y=3x+1   0=<x<5
y=1 at x=0       y=16 at x=5
so 1=<f(x)<16

quadratic functions:
the previous method cannot be used because it is in a shape of a curve and it has a max or min value

eg f(x)=y=x^2  + 2          -2<x<4

if u solve it in the prev. method ----> y=6    y=18  and will consequently write that 6<f(x)<18 which is not right

u have to determine the turning point of the curve
the most straightforward method is to find dy/dx and equating it to 0 which will give u the value of x to get the max. or min point
in the example dy/dx=2x
0=2x -----> x=0
y=2 at x=0        therefore the right range is 2=<f(x)<18

hope u find that useful

[vanibharutham]

Any other last minute queries?