Answer: The ratio is 2.39, which means that the larger acute angle is 2.39 times the smaller acute angle.
Step-by-step explanation:
I suppose that the "legs" of a triangle rectangle are the cathati.
if L is the length of the shorter leg, 2*L is the length of the longest leg.
Now you can remember the relation:
Tan(a) = (opposite cathetus)/(adjacent cathetus)
Then there is one acute angle calculated as:
Tan(θ) = (shorter leg)/(longer leg)
Tan(φ) = (longer leg)/(shorter leg)
And we want to find the ratio between the measure of the larger acute angle and the smaller acute angle.
Then we need to find θ and φ.
Tan(θ) = L/(2*L)
Tan(θ) = 1/2
θ = Atan(1/2) = 26.57°
Tan(φ) = (2*L)/L
Tan(φ) = 2
φ = Atan(2) = 63.43°
Then the ratio between the larger acute angle and the smaller acute angle is:
R = (63.43°)/(26.57°) = 2.39
This means that the larger acute angle is 2.39 times the smaller acute angle.
If you want to include 0, the overall interval is 115 times 0.01, or 23 times 0.05 or 11.5 times 0.10. The latter might make it harder to plot 1.14, so I'd probably use an interval of 0.05.
Between 6 or 7 and about 25 intervals on a graph's scale are about right. More makes it pretty busy and sometimes difficult to tell which mark is associated with the number. A fewer number is indicated only if there are a fewer number of discrete values that need to be shown to adequately identify the data points.
B = 2A/h
The way you would get this is first by clearing the fraction. So in order to clear the fraction, you have to multiply the equation by the denominator.
2(A = 1/2bh)
2A = 1bh
2A = bh
Then you have to isolate the variable. So you'd divide bh by h in order to get b by itself. So you'd end up with:
2A/h = bh/h
2A/h = b
Solve for the x-max or h :\
plug it in to equation to get k:/
-8+16+3=11=k
solve for a by pluging (0,0)
then a= -11/4
Answer:
2.08
Step-by-step explanation:
