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2 years ago

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6. Sketch the region enclosed by the graphs of x =0, 6y-5x=0 and x+3y=21. Find the area

2 answers:

umka2103 [35]2 years ago

6 0

<span>Given three equations, x=0</span>

<span>6y-5x=0</span>

<span>x+3y=21</span>

<span>
</span>

<span>The lines cross at:</span>

<span>x=0, y=0</span>

<span>x=0, y=7</span>

<span>x=6, y=5</span>

<span>so base of triangle is the y-axis between 0 and 7</span>

<span>height is x=6</span>

<span>area=6*7/2=20.5</span>

<span>
</span>

<span>volume=1/3*(base area)*height</span>

<span>=1/3*(pi*6^2)*7</span>

<span>=84pi</span>

<span>=263.9</span>

Lyrx [107]2 years ago

5 0

X = 0

<span>6y - 5x = 0 so it intersects x=0 at (0,0)
</span>
<span>x + 3y = 21
</span>so it intersect x=0 at (0,7)
multiply by 2
<span>2x + 6y = 42
subtracting 6y - 5x = 0
7x = 42
x = 6
so the third line intersects w second line at (6,5)

6. area of triangle = 1/2*base*height
= 1/2*7*6
=21

7. volume = volume of 2 cones
cone volume = 1/3*pi*radius^2*height
lower cone: radius=6, height=5
upper cone: radius=6, height=7-5=2
volume = 1/3*pi*6^2*(5+2)
=84pi
=263.89

</span>

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