The correct mixed number for this fraction 11/3

Mathematics

1 answer:

Sloan [31]3 years ago

6 0

11/3 well 3 goes into 11,,, 3 times with a remainder of 2 so it would be 3 and 2/3.

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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! I mark brainliest :3

vladimir1956 [14]

no for a
yes for B
no for c
that what I think It is opinion could be wrong

3 0

3 years ago

Which expression is equivalent to the following complex fraction?

user100 [1]

The equivalent expression of $(1 + \frac{1}{y}) \div (1-\frac 1y)$ is $\frac{y + 1}{y-1}$