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forsale [732]
3 years ago
6

How many solutions are there to the system of equations?

Mathematics
2 answers:
hichkok12 [17]3 years ago
7 0

Answer: The system of equations has NO SOLUTION.

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

y=mx+b

Where "m" is the slope and "b" is the y-intercept.

Given the following system of equations:

\left \{ {{4x-5y=5} \atop {-0.08x+0.10y=0.10}} \right.

Write the first equation and solve for "y" in order to express it in Slope-Intercept form:

4x-5y=5\\\\-5y=-4x+5\\\\y=\frac{-4x}{-5}+\frac{5}{-5}\\\\y=0.8x-1

You can identify that:

m=0.8\\b=-1

Apply the same procedure with the second equation. Then:

-0.08x+0.10y=0.10\\\\0.10y=0.08x+0.10\\\\y=\frac{0.08x}{0.10} +\frac{0.10}{0.10}\\\\y=0.8x+1

You can identify that:

m=0.8\\b=1

The slopes of both lines are equal, therefore  the lines are parallel and the system has NO SOLUTION.

AfilCa [17]3 years ago
4 0

answer: <u>A</u>

(no solutions)

your welcome

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Answer:

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Step-by-step explanation:

Solution:-

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

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

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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°

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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.

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