Prove that , tan 4x = 4 tan x (1-tan^2 x ) / (1-6 tan^2x + tan^4x).

1 answer:

3 0

Based on your question to prove that both of the equations are true, you must first simplify them. So in accordance with my calculation and practicing step by step procedure i came with a solution and result in an answer of 1-(2tanx/1-tan^2x)^2

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Determine if the following lines are parallel (never intersect), perpendicular (intersect at a 90 degree angle), intersecting (i

mihalych1998 [28]

Answer:

These lines are perpendicular.

Step-by-step explanation:

Put both equations in the slope intercept form of a line.

y = -6x - 8 This is already in the slope intercept form for a line

-x + 6y = 12 Add x to both sides of the equation

6y = x + 12 Divide through the whole equation by 6

y = 1/6x +2

Now we compare the two slopes of -6 and 1/6. They are negative reciprocals of each other. That means that the line are perpendicular.

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2 years ago

the music booster sell 322 music buttons and raise $483 for the music departemen how much does each button cost

nirvana33 [79]

<span>483 / 322 = 1.5. $1.5 is the cost for one button. Hope this helps. :)</span>

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3 years ago

Read 2 more answers

A sample of tritium-3 decayed to 94.5% of its original amount after a year.

Zolol [24]

(a) If $y(t)$ is the mass after $t$ days and $y(0) = A$ then $y(t) = Ae^{kt}.$

$y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945$

$\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\ ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}$

(b)
$Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}$