VCE Maths questions help (1 Viewer)

[boredsatan]

would it be 18.4349488229 degrees?

anyone?

 

[pikachu975]

would it be 18.4349488229 degrees?

No it'd be 71.57 degrees

 

[InteGrand]

would it be 18.4349488229 degrees?

anyone?

The slope m is 3, not 1/3.

 

[boredsatan]

The slope m is 3, not 1/3.

would the equation y = 1/3 x + 5/3 have an angle to the positive x axis of 18.4349488229

 

[boredsatan]

would the equation y = 1/3 x + 5/3 have an angle to the positive x axis of 18.4349488229

anyone?

 

[Squar3root]

would the equation y = 1/3 x + 5/3 have an angle to the positive x axis of 18.4349488229

yes because the gradient of the line is 1/3 so the gradient is tan(1/3) = 18.4 deg

 

[boredsatan]

y = 3/(x-1)+2
the inverse would have a domain of R\{2}?

 

[InteGrand]

y = 3/(x-1)+2
the inverse would have a domain of R\{2}?

Yes, the inverse's domain will be the range of the original function.

 

[boredsatan]

y = 4x^2 - x - 5
find f(x) < 0

 

[Squar3root]

y = 4x^2 - x - 5
find f(x) < 0

factorise it
draw the graph
then you can see were f(x) < 0

 

[boredsatan]

factorise it
draw the graph
then you can see were f(x) < 0

(x+1)(4x-5)
x = -1, 5/4,
i'm confused how to do it after this step

 

[InteGrand]

(x+1)(4x-5)
x = -1, 5/4,
i'm confused how to do it after this step

Sketch the graph.

 

[boredsatan]

Sketch the graph.

would the answer be all the values of x below the x axis?

 

[Squar3root]

would the answer be all the values of x below the x axis?

yes because that is where the y values are negative (which is what you want)

 

[boredsatan]

yes because that is where the y values are negative (which is what you want)

but if it's asking f(x) < 0, why are we meant to find the negative y values?

 

[InteGrand]

but if it's asking f(x) < 0, why are we meant to find the negative y values?

Because the graph was of y = f(x).

The solution will be the set of x values for which the curve is below the x-axis.

 

[boredsatan]

yes because that is where the y values are negative (which is what you want)

Are you doing year 12 this year?
Just wondering

 

[Squar3root]

Are you doing year 12 this year?
Just wondering

yeah why

try not to derail

you can PM me

 

[boredsatan]

What are the solutions to the cubic inequation (x-a)^2 (x+b) > 0, where a and b are both positive constraints

 

[boredsatan]

What are the solutions to the cubic inequation (x-a)^2 (x+b) > 0, where a and b are both positive constraints

anyone?