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Semenov [28]
3 years ago
12

Two cylindrical containers, C and D, with different capacities are shown below. What is the total volume of liquid held in these

containers if half of container C is full and one third of container D is full?

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
6 0

Answer:

Option D. (x + 4)(x + 1)

Step-by-step explanation:

From the question given above, the following data were obtained:

C = (6x + 2) L

D = (3x² + 6x + 9) L

Also, we were told that half of container C is full and one third of container D is full. Thus the volume of liquid in each container can be obtained as follow:

Volume in C = ½C

Volume in C = ½(6x + 2)

Volume in C = (3x + 1) L

Volume in D = ⅓D

Volume in D = ⅓(3x² + 6x + 9)

Volume in D = (x² + 2x + 3) L

Finally, we shall determine the total volume of liquid in the two containers. This can be obtained as follow:

Volume in C = (3x + 1) L

Volume in D = (x² + 2x + 3) L

Total volume =?

Total volume = Volume in C + Volume in D

Total volume = (3x + 1) + (x² + 2x + 3)

= 3x + 1 + x² + 2x + 3

= x² + 5x + 4

Factorise

x² + 5x + 4

x² + x + 4x + 4

x(x + 1) + 4(x + 1)

(x + 4)(x + 1)

Thus, the total volume of liquid in the two containers is (x + 4)(x + 1) L.

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3 years ago
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zmey [24]

Answer:

<h2>140°</h2>

solution,

Let <DEF= x°

<CDE= 2x°

<BAF= 90°

The sum of interior angle of hexagon= 720°

<A + <B + <C + <D + <E +<F = 720°

90 + 145 + 115 + 2x + x + 160 = 720 \\ 510 + 3x = 720 \\3x + 510 = 720 \\ 3x + 510 - 510 = 720 - 510 \\ 3x = 210 \\  \frac{3x}{x}  =  \frac{210}{3}  \\ x = 70 \: degree

Again,

< \: cde = 2x \\  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  = 2 \times 70 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 140 \: degree

Hope this helps...

Good luck on your assignment..

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3 years ago
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