I COULD REALLY USE THE HELP Complete the following problems. Show all your work. Round all side lengths to the nearest tenth and all degrees to the nearest degree. When applicable, leave your answers in radical form.
1. Name the following sides of the triangle.

  see attachment

Step-by-step explanation:

When there are quite a few similar calculations to do, it can work well to let a spreadsheet do them.

5. The Pythagorean theorem applies for finding the unknown side:

  DF = √(FE^2 -DE^2) = 3.0

The angles can be found using any of the trig functions. Since we were given the length of a side and the hypotenuse, we choose to use the sine function:

  Sin = Opposite/Hypotenuse

  opposite angle = arcsin(Opposite/Hypotenuse)

In the spreadsheet, Leg 1 is the given leg, and Leg 2 is the computed leg. The associated Angle 1 and Angle 2 are opposite those legs:

 ∠F = 59°

  ∠E = 31°

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6. In this section, there are two kinds of problems. One has the hypotenuse and an angle given (HA problems). The other has a leg and its opposite angle given (LA problems).

HA problems (6c, 6e, 6f)

We call the given angle "Angle 1" and compute the other angle as its complement. The side opposite Angle 1 is Leg 1. It is computed using the sine relation above.

  Opposite = Hypotenuse×Sin

  x = 16·sin(60°) = 13.9 . . . . . . . . . . for problem 6c

Then the Adjacent leg is found using the cosine relation:

  Cos = Adjacent/Hypotenuse

  Adjacent = Hypotenuse×Cos

  y = 16·cos(60°) = 8.0 . . . . . . . . . . for problem 6c

As in problem 5, the missing angle is computed as the complement of the given angle.

  missing angle = 90° -60° = 30° . . . . . . for problem 6c

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LA problems (6a, 6b, 6d)

The calculations for these problems are similar, except that we compute the hypotenuse first using the sine relation.

  Hypotenuse = Opposite/Sin

  u = 2/sin(30°) = 4.0 . . . . . . . for problem 6a

As in the above problems, we computed the missing leg using the cosine function and the hypotenuse.

  v = 4.0·cos(30°) = 3.5 . . . . . for problem 6a

And the missing angle is computed the same as for HA problems.

  missing angle = 90° -30° = 60° . . . . for problem 6a

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The computations are done in the spreadsheet using the formulas described above. The matchup of Leg1, Leg2, Angle1, and Angle2 to the problem variables and vertices is left to the student. As we said, Leg1 and Angle1 are opposite each other, and correspond to the given leg and/or angle.

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Comment on radians

When using sine and cosine functions in a spreadsheet, one needs to be aware that their arguments must be in radians. Fortunately, the spreadsheet provides functions for converting angles back and forth between degrees and radians. Similarly, the arcsin function used in problem 5 gives its result in radians, so requires conversion to degrees.

theres no picture: /

step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: