1/10 x 90 I think that's the answer
Answer:
about 2949 feet
Step-by-step explanation:
The geometry of the situation can be modeled by a right triangle. The height of the cliff can be taken to be the side opposite the given angle, and the distance to the coyote will be the side adjacent to the given angle. The relation between these values is the trig function ...
Tan = Opposite/Adjacent
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<h3>setup</h3>
Filling in the known values, we have ...
tan(6°) = (310 ft)/(distance to coyote)
<h3>solution</h3>
Multiplying by (distance to coyote)/tan(6°) gives ...
distance to coyote = (310 ft)/tan(6°) ≈ 310/0.105104 ft
distance to coyote ≈ 2949.453 ft
The coyote is about 2949 feet from the base of the cliff.
To simplify the expression it would be
Steve set the equation equal to 50 which is the entire length of JL. instead he should've set the equation equal to 25, since it's half of 50 and would represent the midpoint. he should've said 2x + 5 = 25
4 to 7 = 4/7
5/8 is not.
12/20 = 6/10 = 3/5 . It is not equivalent.
8:14 = 4 : 7 . It is equivalent
5 to 10 = 1 to 2. It is not.
16/28 = 8/14 = 4/7. It is equivalent.
9 to 16 = 9 to 16. It is not equivalent.
So only 8:14 and 16/28 are equivalent to 4 to 7.