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nordsb [41]
3 years ago
13

Graph parallelogram ABCD on the graph

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
3 0

Answer: 25 square units

Step-by-step explanation:

We mark the points, A(2,0), B(7,0), C(10,3), D (5,3). on a graph and then joined them to make parallelogram ABCD as provided in the attachment.

Area of parallelogram  = Base x corresponding height

From the figure, base AB = 7 - 2 units = 5 units

corresponding height: h= 5 units

Now , Area of parallelogram  ABCD = base AB x corresponding height

= 5 x 5 square units

= 25 square units

Hence,  the area of parallelogram  ABCD is 25 square units .

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Use the disk method or the shell method to find the volumes of the solids generated by revolving the region bounded by the graph
diamong [38]

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c) 32/5 π

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Step-by-step explanation:

Given lines :  y = √x, y = 2, x = 0.

<u>a) The x-axis </u>

using the shell method

y = √x = , x = y^2

h = y^2 , p = y

vol = ( 2π ) \int\limits^2_0 {ph} \, dy

     = ( 2\pi ) \int\limits^2_0 {y.y^2} \, dy  

∴ Vol = 8π

<u>b) The line y = 2  ( using the shell method )</u>

p = 2 - y

h = y^2

vol = ( 2π ) \int\limits^2_0 {ph} \, dy

     = ( 2\pi ) \int\limits^2_0 {(2-y).y^2} \, dy

     = ( 2π ) * [ 2/3 * y^3  - y^4 / 4 ] ²₀

∴ Vol  = 8/3 π

<u>c) The y-axis  ( using shell method )</u>

h = 2-y  = h = 2 - √x

p = x

vol = (2\pi ) \int\limits^4_0 {ph} \, dx

     = (2\pi ) \int\limits^4_0 {x(2-\sqrt{x}  ) } \, dx

     = ( 2π ) [x^2 - 2/5*x^5/2 ]⁴₀

vol = ( 2π ) ( 16/5 ) = 32/5 π

<u>d) The line x = -1    (using shell method )</u>

p = 1 + x

h = 2√x

vol = (2\pi ) \int\limits^4_0 {ph} \, dx

Hence   vol = 176/15 π

attached below is the graphical representation of P and h

3 0
3 years ago
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