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:
brainly.com/question/14033725
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Answer:
a. y = (3/5)× x +1
Step-by-step explanation:
y= mx+c
m = (13-4)/(20-5) = 9/15 = 3/5
c= 1
so, y= mx+c becomes y= (3/5) × x +1
Answer:
Vf = 118.365 m3
Step-by-step explanation:
Vcil = Pi*[(20)^2]*(100=
Vcil = 125.600 m3
Vesf = (4/3)*Pi*(12)^3
Vesf = 7235 m3
Then
Vf = Vci - Vesf
Vf = 125.600 - 7235
Vf = 118365 m3
Best regards
Answer:
4 * (x-9)=-30
not sure if u want it solved but if u do its x=6/4
Step-by-step explanation:
B because both lines a and b are parallel which makes 1 and 9 congruent