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Rzqust [24]
3 years ago
15

Please help me with these geometry questions.

Mathematics
1 answer:
sashaice [31]3 years ago
7 0

Answer:

1)2(lb+bh+hl)

=2(17*12+12*8+8*17)

=2(204+96+136)

=2*436

=872 in^2

2)2*pi*r(r+h)

=2*3.14*14(14+29)

=6.28*14(43)

=6.28*14*43

=3780.56 mm^2

3)A=11*4.3+3*6 +3*8+11*3(Area of each plane figure -triangle and rectangle)

=47.3+18+24+33

=122.3cm^3

4)h*(sum of II sides)+14*8+19*8+14*8+25(Area of plane surface-trapezium,rectangle,square)

=12.1(19+5)+112+152+112+25

=12.1*24+401

=290.4+401

=691.4ft^2

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Anon25 [30]

Answer:

0.6

Step-by-step explanation:

The x increases by 5, while y increases by 3

Lets use the method, which is \frac{rise}{run}.

Keep in mind that x goes horizontally along the axis, while y goes vertically on the axis.

So plug the numbers in.

\frac{3}{5} = 0.6

Hence, the constant proportionality is 0.6.

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Answer:

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4 years ago
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3 years ago
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A hollow toy, with the dimensions shown in the figure, is to be stuffed with rigid foam. What is the maximum amount of foam that
Fudgin [204]

Option B: 67.02 cubic inches

<h2>Given that:</h2>

The diagram consists of two parts:

  • First a cone on the top with 12 inches height and 4 inches diameter.
  • Second a semi sphere with diameter of 4 inches.

<h2>Calculations:</h2>

The amount of foams that can be stuffed with foam is total volume of both the containers.

Volume of the given cone of 12 inches height and 4 inches diameter can be given as:

\dfrac{1}{3} \times \pi \times 2^2 \times 12\\= 50.265 \: \rm inch ^3

Volume of the semi sphere of 4 inches diameter can be given as:

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Thus total volume is 50.265 + 16.755 = 67.02 cubic inches.

Thus option B : 67.02 cubic inches is the needed volume.

Learn more here:

brainly.com/question/1315822

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