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2 years ago

Elijah and Jonathan play on the same soccer team. They have played 3 of their 15,

Mathematics

1 answer:

Lynna [10]2 years ago

5 0

Answer:

Only Elijah's model is correct

Step-by-step explanation:

The data given in the question tells us they have 12 games left on their soccer team. Each one of them tried to simulate the fact by creating some model which look like a balance between quantities.

Elijah placed 3 cubes of value 1 and a cube of value x on one side of a balance. On the other side, he placed 15 cubes of value 1. He was obviously modeling the fact that 15 cubes (games due to play in our case) should be equal to 3 cubes (games already played) plus the x numbers left to play

This model if perfect, since the only way to equilibrate the balance is setting x to 12, the games left to play

Jonathan used a table with 3 x's in a row and a 15 in the second row, trying to model the same situation. To our interpretation, this table doesn't show the number of games left to play. If we equate 3x = 15, we get x=5 which has nothing to do with the situation explained in the question, so this model is not correct.

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$\tan A=\dfrac{\sin A}{\sin C}$

By the definition,

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Now

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By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.$

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Option B is false.

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By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

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Option C is false.

$\cos A=\tan C.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

As you can see $\cos A\neq \tan C$ and option D is not true.

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By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.$

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$\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.$

This option is false.

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Answer:

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Step-by-step explanation:

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