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ddd [48]
3 years ago
5

What is the area of the parallelogram with vertices at (0,1), (-1,2), (5,4) and (6,3)?

Mathematics
1 answer:
marin [14]3 years ago
7 0
: <span>A=<span>√265</span></span><span>.     hope this helps jngfjndsghndfhcdghhsdfhf</span>
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sergejj [24]

Answer:

Step-by-step explanation:

We have the equations

4x + 3y = 18   where x = the side of the square and y = the side of the triangle

For the areas:

A = x^2 + √3y/2* y/2

A = x^2  + √3y^2/4

From the first equation x = (18 - 3y)/4

So substituting in the area equation:

A = [ (18 - 3y)/4]^2 + √3y^2/4

A = (18 - 3y)^2 / 16 + √3y^2/4

Now for maximum / minimum area the derivative = 0 so we have

A' = 1/16 * 2(18 - 3y) * -3 + 1/4 * 2√3 y = 0

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-27/4 + 9y/8 + √3y /2 = 0

-54 + 9y + 4√3y = 0

y = 54 / 15.93

= 3.39 metres

So x = (18-3(3.39) / 4 = 1.96.

This is a minimum value for x.

So the total length of wire the square  for minimum  total area is 4 * 1.96

= 7.84 m

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3 years ago
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Step-by-step explanation:


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3 years ago
A) x2 when x = 5<br> b) y2 when y = -2<br> c) z2 when z = 0.5
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Answer:

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andrew-mc [135]

Answer:

x = 12

Step-by-step explanation:

The sum of the two smaller angles is 140

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Subtract 80 from each side

80+5x-80 =140-80

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Divide each side by 5

5x/5 = 60/5

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Answer:

The y-intercept is 33ft.

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