What is 3x+4-(-6) and what do I need to do

Mathematics

1 answer:

34kurt3 years ago

6 0

Answer:

3x + 10

I just combined like terms.

Hope this helps!

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Find the exact values of sin2 θ for cos θ = 3/18 on the interval 0° ≤ θ ≤ 90°

mote1985 [20]

Answer:

$sin(2\theta)=\frac{\sqrt{35} }{18}$

Step-by-step explanation:

Recall the formula for the sine of the double angle:

$sin(2\theta)=2*sin(\theta)*cos(\theta)$

we know that $cos(\theta)=\frac{3}{18}$ , and that $\theta$ is in the interval between 0 and 90 degrees, where both the functions sine and cosine are non-negative numbers. Based on such, we can find using the Pythagorean trigonometric property that relates sine and cosine of the same angle, what $sin(\theta)$ is:

$cos^2(\theta)+sin^2(\theta)=1\\sin^2(\theta)=1-cos^2(\theta)\\sin(\theta)=\sqrt{1-cos^2(\theta)} \\sin(\theta)=\sqrt{1-(\frac{3}{18} )^2}\\sin(\theta)=\sqrt{1-\frac{9}{324} }\\sin(\theta)=\sqrt{\frac{324-9}{324} }\\sin(\theta)=\sqrt{\frac{315}{324} }\\\\sin(\theta)=\frac{3}{18}\sqrt{35 }$

With this information, we can now complete the value of the sine of the double angle requested:

$sin(2\theta)=2*sin(\theta)*cos(\theta)\\sin(2\theta)=2*\frac{3}{18} \,\sqrt{35} \,\frac{3}{18}\\sin(2\theta)=\frac{2*3*3}{18*18}\,\sqrt{35} \\sin(2\theta)=\frac{\sqrt{35} }{18}$