The inclination to the nearest tenth of a degree exists 0.24146
<h3>What is the inclination to the nearest tenth of a degree?</h3>
The given scenario includes a right-angled triangle where the length of the ramp exists hypotenuse and the rise of ramp exists the perpendicular.
Given: H = 4.6 m and P = 1.1 m
We have to use the trigonometric ratios to find the angle. The ratio that has to be used should involve both perpendicular and hypotenuse
Let x be the angle then
sin x = P/H
sin x = 1.1/4.6
sin x = 0.23913
= 0.24146
The inclination to the nearest tenth of a degree exists 0.24146
To learn more about trigonometric ratios refer to:
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Answer:
its 36 gallons
Step-by-step explanation:
You are taking off 8 each 2 at x, so it would be four for each 1 at x. therefore 2 is 9, 1 is 13 and...
the value of y is 17 when x = 0
sorry if i explained that bad
Answer: The central angle of the arc is 162 degrees.
Step-by-step explanation: The information available are as follows;
Circumference of the circle equals 10. Length of an arc equals 9/2. The circumference of a circle is given as;
Circumference = 2Pi x r
That means 2Pi x r = 10.
Also the length of an arc along the same circle is 9/2. Length of an arc is calculated as;
Length of arc = (X/360) x 2Pi x r
Where X is the central angle of the arc
That means;
9/2 = (X/360) x 2Pi x r
We can now substitute for the known values as follows
Length of an arc = (X/360) x 2Pi x r
9/2 = (X/360) x 10
9/2 = 10X/360
By cross multiplication we now have
(9 x 360)/(2 x 10) = X
3240/20 = X
162 = X
The angle at the center of the arc is 162 degrees.
This is a great question, but it's also a very broad one. Please find and post one or two actual rational expressions, so we can get started on specifics of how to find vertical and horiz. asymptotes.
In the case of vert. asy.: Set the denom. = to 0 and solve for x. Any real x values that result indicate the location(s) of vertical asymptotes.
