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Mnenie [13.5K]
4 years ago
6

Least to greatest 0.6,1/2, 2/3, 0.39

Mathematics
1 answer:
lorasvet [3.4K]4 years ago
6 0

Answer:

0.39, 1/2, 0.6, and 2/3

Step-by-step explanation:

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What is the answer of 24-4m=6(7m-4)+2
otez555 [7]
24-4m=42m-24+2
24+24-2=42m+4m
46=46m
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6 0
3 years ago
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Help me understand please ;-;
Novosadov [1.4K]
Standard form for first one is
ax+by=c

6y-12=-3x
add 3x both sides
3x+6y-12=0
add 12
3x+6y=12


second one
distribute (don't forget double negative y)
10-2x+2y=3x+1
minus 3x both sides
10-5x+2y=1
minus 10 both sides
-5x+2y=-9



the equations are
3x+6y=12
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4 years ago
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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
4 years ago
Reduce fraction 27/36 lowest term
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The lowest term is 3
                            ---
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<span>0.888 = 888 / 1000

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