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Korvikt [17]
3 years ago
9

Solve image below:

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
4 0

Answer: x=\frac{6}{-3y+8}

Step-by-Step Explanation:

Let's solve for x.

\frac{x-2}{3y-5} =\frac{x}{3}

Step 1: Multiply both sides by 3y-5.

x-2=\frac{3xy-5x}{3}

Step 2: Multiply both sides by 3.

3x-6=3xy-5x

Step 3: Add -3xy to both sides.

3x-6+-3xy=3xy-5x+-3xy

-3xy+3x-6=-5x

Step 4: Add 5x to both sides.

-3xy+3x-6+5x=-5x+5x

-3xy+8x-6=0

Step 5: Add 6 to both sides.

-3xy+8x-6+6=0+6

-3xy+8x=6

Step 6: Factor out variable x.

x(-3y+8)=6

Step 7: Divide both sides by -3y+8.

\frac{x(-3y+8)}{-3y+8}=\frac{6}{y-3y+8}

x=\frac{6}{-3y+8}

Answer:

x=\frac{6}{-3y+8}

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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En un triángulo rectángulo A es un ángulo agudo y Sen A = 4/5 ¿Cuál será el valor de Tan A?
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Answer:

\displaystyle \tan A=\frac{4}{3}

Step-by-step explanation:

<u>Funciones Trigonométricas</u>

La identidad principal en trigonometría es:

sen^2A+cos^2A=1

Si sabemos que A es un ángulo agudo (que mide menos de 90°), su seno y coseno son positivos.

Dado que Sen A = 4/5, calculamos el coseno:

cos^2A=1-sen^2A

Sustituyendo:

\displaystyle cos^2A=1-\left(\frac{4}{5}\right)^2

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\displaystyle cos^2A=\frac{25-16}{25}

\displaystyle cos^2A=\frac{9}{25}

Tomando raíz cuadrada:

\displaystyle cos\ A=\sqrt{\frac{9}{25}}=\frac{3}{5}

La tangente se define como:

\displaystyle \tan A=\frac{sen\ A}{cos\ A}

Substituyendo:

\displaystyle \tan A=\frac{\frac{4}{5}}{\frac{3}{5}}

\displaystyle \tan A=\frac{4}{3}

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Step-by-step explanation:

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