Salina, Tamara, and Uma are working on their homework together. Salina tells her friends that she got θ=52∘θ=52∘ for the inverse cosine problem she is working on and asks if they got the same answer. Tamara says she got θ=128∘θ=128 ∘ and Uma volunteers her answer of θ=308∘θ=308∘. Is it possible that these are all solutions to the same problem? Justify your answer.

The values reported by Uma, Salina, and Tamara are all possible values for inverse cosine

Step-by-step explanation:

Solution:-

- The solution to the inverse "cosine" problem would have a general form:

                                    θ = cos^-1 ( a )

Where,  a: Any arbitrary constant, - 1 < a < 1

- The value of θ = 52° was reported by Salina suggests that the answer lies in the first quadrant of a cartesian plane where ( sin (θ) , cos (θ) , tan (θ) ) have positive values for "a".

Hence,                            0 < a < 1 , θ = 52°

- The value of θ = 128° was reported by Tamara suggests that the answer lies in the second quadrant of a cartesian plane where ( sin (θ) ) have positive values for "a" and (cos (θ) , tan (θ) have negative values for "a". So for cos (128):

Hence,                           -1 < a < 0 , θ = 128°

- The value of θ = 308° was reported by Uma suggests that the answer lies in the fourth quadrant of a cartesian plane where ( cos (θ) ) have positive values for "a" and (sin (θ) , tan (θ) have negative values for "a". So for cos (308):

Hence,                           0 < a < 1 , θ = 308°

- The angle θ reported by Uma and Salina are similar solution because of property law of complementary angles:

                                     cos (θ) = cos ( 360 - θ )

Where, θ = 52°,            cos (52°) = cos( 308°)  .. Uma and Salina conform

However,                      cos ( 180 - θ ) = - cos (θ)  

                                     cos(128) = - cos ( 52 ) Uma and Tamara conform.

answer: i don't understand it

step-by-step explanation: sorry

answer:

total cost for tiles and paints is $924.  

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

given: 1m squared floor tiles costs $3. so,

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

therefore, the combined total cost of tiles and paint is $924.