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galben [10]
2 years ago
6

Solve for X Log x - log 3 = log(x-2) Please please help

Mathematics
1 answer:
andre [41]2 years ago
8 0

Step-by-step explanation:

log \frac{a}{b}  = loga - logb

logx - log3 = log(x-2)

log( \frac{x}{3} ) = log(x - 2)

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

x = 3(x - 2)

x = .......

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What is the mean of 65.9, 80.8, 60.8, 46.5, and 64.3
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63.66

Step-by-step explanation:

Mean=(65.9+80.8+60.8+46.5+64.3)/5

Mean=63.66

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3 years ago
Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 4 units, b = 5 units, and c = 9
taurus [48]
From the figure, let the distance of point P from point A on line segment AB be x and let the angle opposite side a be M and the angle opposite side c be N.

Using pythagoras theorem,
\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)
and
\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)

Angle θ is given by
\theta=180-M-N \\  \\ =180-\tan^{-1}\left(\frac{a}{b-x}\right)-\tan^{-1}\left(\frac{c}{x}\right)

Given that a = 4 units, b = 5 units, and c = 9 units, thus
\theta=180-\tan^{-1}\left(\frac{4}{5-x}\right)-\tan^{-1}\left(\frac{9}{x}\right)

To maximixe angle θ, the differentiation of <span>θ with respect to x must be equal to zero.
i.e.
\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\  \\ -4(x^2+81)+9(x^2-10x+41)=0 \\  \\ -4x^2-324+9x^2-90x+369=0 \\  \\ 5x^2-90x+45=0 \\  \\ x^2-18x+9=0 \\  \\ x=9\pm6 \sqrt{2}

Given that x is a point on line segment AB, this means that x is a positive number less than 5.

Thus
x=9-6 \sqrt{2}=0.5147

Therefore, The distance from A of point P, so that </span>angle θ is maximum is 0.51 to two decimal places.
6 0
3 years ago
PLEASE HELP ME I GIVE THE BRAINLIEST
Tems11 [23]

Answer:

what grade you in

Step-by-step explanation:

giuhyiufhyeiufheoirgjuoeirgjuer

7 0
3 years ago
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