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ololo11 [35]
3 years ago
13

Find the perimeter of the parallelogram with these vertices.

Mathematics
1 answer:
blondinia [14]3 years ago
4 0

Check the picture below.

so we can say that two sides are "4" each in length, since opposite sides are equal, let's find how long the slanted sides are.

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[3 - (-4)]^2 + [5 - 2]^2}\implies d=\sqrt{(3+4)^2+3^2} \\\\\\ d=\sqrt{49+9}\implies d=\sqrt{58} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{4~~ + ~~4~~ + ~~\sqrt{58}~~ + ~~\sqrt{58}\implies 8+2\sqrt{58}}

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A trough is 10 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1
Anna11 [10]

Answer:

7.5 ft³/min

Step-by-step explanation:

Let x be the depth below the surface of the water. The height, h of the water is thus h = 10 - x.

Now, the volume of water V = Ah where A = area of isosceles base of trough = 1/2bh' where b = base of triangle = 4 ft and h' = height of triangle = 1 ft. So, A = 1/2 × 4 ft × 1 ft = 2 ft²

So, V = Ah = 2h = 2(10 - x)

The rate of change of volume is thus

dV/dt = d[2(10 - x)]/dt = -2dx/dt

Since dV/dt = 15 ft³/min,

dx/dt = -(dV/dt)/2 = -15 ft³/min ÷ 2 = -7.5 ft³/min

Since the height of the water is h = 10 - x, the rate at which the water level is rising is dh/dt = d[10 - x]/dt

= -dx/dt

= -(-7.5 ft³/min)

= 7.5 ft³/min

And the height at this point when x = 8 inches = 8 in × 1 ft/12 in  = 0.67 ft is h = (10 - 0.67) ft = 9.33 ft

7 0
3 years ago
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amid [387]
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Thepotemich [5.8K]

Answer:

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Answer:

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3 years ago
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