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3 years ago

12

Trigonometry problem. Please show your solution and explain it. thank you !!

1 answer:

WINSTONCH [101]3 years ago

7 0

Cute one!
<span>
</span>Summarizing:
<span>sec(acot(tan(asin(sin(pi/3)))) .... use asin(sin(x))=x
</span>=sec(acot(tan(pi/3)))
=sec(acot(sqrt(3))) ......... use acot(x)=atan(1/x)
=sec(atan(1/sqrt(3)))
=sec(atan(sqrt(3)/3)) .... evaluate atan(sqrt(3)/3), use unit circle
=sec(pi/6)
=1/cos(pi/6)...... evaluate cos(pi/6), use unit circle
=1/(sqrt(3)/2)
=2/sqrt(3) .... now rationalize
=2sqrt(3)/3

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4 years ago

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Answer:

D: As the x-values go to negative infinity, the function's values go to positive infinity.

Step-by-step explanation:

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When x-values go to positive infinity, we see that f(x) values also go to positive infinity.

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4 years ago

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Find the zeros of y = x2 – 6x – 4 by completing the square.

katrin [286]

Answer:

The solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$

Step-by-step explanation:

Given the function

$y\:=\:x^2\:-\:6x\:-\:4$

substitute y = 0 in the equation to determine the zeros

$0\:=\:x^2\:-\:6x\:-\:4$

Switch sides

$x^2-6x-4=0$

Add 4 to both sides

$x^2-6x-4+4=0+4$

Simplify

$x^2-6x=4$

Rewrite in the form (x+a)² = b

But, in order to rewrite in the form x²+2ax+a²

Solve for 'a'

2ax = -6x

a = -3

so add a² = (-3)² to both sides

$x^2-6x+\left(-3\right)^2=4+\left(-3\right)^2$

$x^2-6x+\left(-3\right)^2=13$

Apply perfect square formula: (a-b)² = a²-2ab+b²

$\left(x-3\right)^2=13$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x-3=\sqrt{13}$

Add 3 to both sides

$x-3+3=\sqrt{13}+3$

Simplify

$x=\sqrt{13}+3$

now solving

$x-3=-\sqrt{13}$

Add 3 to both sides

$x-3+3=-\sqrt{13}+3$

Simplify

$x=-\sqrt{13}+3$

Thus, the solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$

4 0

3 years ago