How do you calculate the forest area?
The solutions state g(x)=1/3x-2. But how do you get the gradient m=1/3x?
The solutions state g(x)=1/3x-2. But how do you get the gradient m=1/3x?
Hello, a quick question about the distributive law. One of my flashcards says the following: However, this example refers to a single operator within the parentheses. What happens if we have a term inside the parentheses? Example: (9 + 9 * 9) May I continue to apply the distributive law to the first factor, i.e….
Can ax also be considered a factor form?
Hello, I'm currently writing my VWA and I haven't made much progress. So far, we haven't had a group meeting with our supervisor. Nobody knows what's going on. She keeps canceling everyone, and I don't know anything about it either. I don't even know how to start or what it should contain. My topic is…
In this task I first calculated k with k/l = b/c then m and a Unfortunately that was wrong… but I can’t find the mistake Could someone please tell me where I made a mistake?
y = mx + b
-2 = 0m + b, i.e. b = -2
0=6m-2, i.e. 6m=2, i.e. m=2/6=1/3
Most Northern point of the forest: f(x) = 0 set (maximum).
The forest area it the triangular area plus the integral of function f(x), i.e. the given root function from 0 to 6.
Detection of the stam function by deriving F(x) with product rule and optionally chain rule.
The highway has an angle of 90°, the river has the derivation = slope of f(x) at x = 0.
Dfference of slopes form!
Calculate turning point of f(x). How large is the pitch of the fuction f(x) at the turning point? Set the straight line of the tangent.
Set straight line for the forest path. Your gradient is the negative reciprocal of the tangent pitch!
The point of intersection P is finished.
This is the equation of the footpath.
At 6 km in the x-direction y rises by 2 km, the slope
is therefore 2/6 = 1/3.