Answer:
hihi
Step-by-step explanation:
Answer:
should be 17
Step-by-step explanation:
Answer:
Ignacio can make 5 clockwise rotations.
Step-by-step explanation:
Given that the Ignacio's legs are already at a height of 49.3cm and each rotation of the chair knob raises his legs another 4.8cm, we can set up an inequality to determine the number of rotations Ignacio could make without his legs touching the desk, which is at a height of 74.5cm:
4.8r + 49.3 < 74.5 where 'r' is equal to the number of rotations
The sum of the Ignacio's original leg height plus the amount of height increased from the rotations of the know must be less than 74.5 in order for his legs not to touch. Now, solve for 'r':
Subtract 49.3 from both sides: 4.8r + 49.3 - 49.3 < 74.5 - 49.3 or 4.8r < 25.2
Divide 4.8 from both sides: 4.8r/4.8 < 25.2/4.8 or r < 5.25
Since the number of rotations must be less than 5.25, he can make 5 complete rotations.
Answer: sin u = -5/13 and cos v = -15/17
Step-by-step explanation:
The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.
So for u: (-5)^2 + adj^2 = 13^2, so with what you gave us (Quad 3),
==> adj of u = -12 therefore cos u = -12/13
Same argument for v: adj = -15,
opp^2 + (-15)^2 = 17^2 ==> opp = -8 therefore sin v = -8/17
The cosine rule for cos (u + v) = (cos u)(cos v) - (sin u)(sin v) and now we substitute: cos (u + v) = (-12/13)(-15/17) - (-5/13)(-8/17)
I am too lazy to do the remaining arithmetic, but I think we have created a way to approach all of the similar problems.
