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3 years ago

In a right triangle, Sin (30 + x) º = Cos (3x) º. What is the value of x?

Mathematics

1 answer:

pogonyaev3 years ago

5 0

Answer:

$x=15^\circ$

Step-by-step explanation:

In a right triangle:

$\sin \theta =cos(90^\circ-\theta) $ (Complementary angles)$

Therefore, given:

$Sin (30 + x) \º = Cos (3x) \º\\30+x+3x=90^\circ\\30+4x=90^\circ\\4x=90^\circ-30^\circ\\4x=60^\circ\\$Divide both sides by 4\\x=15^\circ$

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The segment addition postulate states that if point C is between points A and B, then... ANSWER ASAP

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<u>Given</u>:

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Read 2 more answers

Which equation is y = â€""6x2 3x 2 rewritten in vertex form?.

Ber [7]

The vertex form needed is :

$-6(x-0.25)^2 +2.375$

Given the equation:

$-6x^2 + 3x + 2\\$

The vertex form of $f(x) = ax^2 + b(x) + c$ is given by:

$f(x) = a(x-h)^2 + k$

Here, (h,k) is the vertex of the parabola equation given.

Conversion will go like this:

$-6x^2 + 3x + 2\\= -6(x^2 - 0.5x) + 2\\= -6(x^2 -0.5x + 0.0625 - 0.0625) + 2\\= -6(x^2 -0.5x + 0.0625) + 2 +0.375\\= -6(x-0.25)^2 + 2.375\\\\$

This is the resultant vertex form.

Thus the vertex form needed is :

$-6(x-0.25)^2 +2.375$

Learn more here:

brainly.com/question/15165354

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