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trasher [3.6K]
3 years ago
6

A certain pipe can be used to fill a pool with water in 2 hours. A second pipe can be used to fill the pool in 4 hours. How long

does it take to fill the pool when both pipes are used at the same time?​
Mathematics
1 answer:
Kryger [21]3 years ago
6 0

Answer:

1 1/3 hours

Step-by-step explanation:

Pipe 1 alone:

fills the pool in 2 hours

in 1 hour, it fills 1/2 of the pool

Pipe 2 alone:

fills the pool in 4 hours

in 1 hour, it fills 1/4 of the pool

Pipe 1 and Pipe 2 working together:

fill the pool in x hours

in 1 hour, they fill 1/x of the pool

Working together, in 1 hour the two pipes fill 1/2 + 1/4 of the pool.

Working together, in 1 hour the two pipes fill 1/x of the pool.

Therefore,

1/2 + 1/4 = 1/x

2/4 + 1/4 = 1/x

3/4 = 1/x

x = 4/3

x = 1 1/3

Answer: 1 1/3 hours

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A bottle of soft drink is two thirds full. Jack drinks one quarter of the remaining soft drink. How full is the bottle now
marusya05 [52]

We know that the bottle of soft drink has a remaining volume which is 2 / 3 of the original.

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<span>Hence the bottle is now one half full</span>

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Which expressions can be used to compute the approximate area of triangle XYZ? Check all that apply.
lesantik [10]

Given:

In the given ΔXYZ,

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Height = 23.1

Also,

Base = 31.9

Height = 15.9

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The area of a triangle is

A = \frac{1}{2} bh

where, b be the base and

h be the height.

Now,

Taking, b = 22 and h = 23.1 we get,

A = \frac{1}{2} (22)(23.1) sq unit

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Taking,

b = 31.9 and h = 15.9 we get,

A = \frac{1}{2} (31.9)(15.9) sq unit

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We can find the approximate area of the triangle by following two expression

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b) \frac{1}{2} (31.9)(15.9)

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point b on the ground is 5 cm from point E at the entrance to Ollie's house. He is 1.8 m tall and is standing at Point D, below
enot [183]

Point B on the ground is 5 cm from point E at the entrance to Ollie's house.

Ollie is at a distance of 2.45 m from the entrance to his house when he first activates the sensor.

The complete question is as follows:

Ollie has installed security lights on the side of his house that is activated by a  sensor. The sensor is located at point C directly above point D. The area covered by the sensor is shown by the shaded region enclosed by triangle ABC. The distance from A to B is 4.5 m, and the distance from B to C is 6m. Angle ACB is 15°.

The objective of this information is:

  • To find angle CAB and;
  • Find the distance Ollie is from the entrance to his house when he first activates the sensor.

The diagrammatic representation of the information given is shown in the image attached below.

Using  cosine rule to determine angle CAB, we have:

\mathbf{\dfrac{AB}{Sin \hat {ACB}} = \dfrac{BC}{Sin \hat {CAB}}= \dfrac{CA}{Sin \hat {ABC}}}

Here:

\mathbf{\dfrac{AB}{Sin \hat {ACB}} = \dfrac{BC}{Sin \hat {CAB}}}

\mathbf{\dfrac{4.5}{Sin \hat {15^0}} = \dfrac{6}{Sin \hat {CAB}}}

\mathbf{Sin \hat {CAB} = \dfrac{Sin 15 \times 6}{4.5}}

\mathbf{Sin \hat {CAB} = \dfrac{0.2588 \times 6}{4.5}}

\mathbf{Sin \hat {CAB} = 0.3451}

∠CAB = Sin⁻¹ (0.3451)

∠CAB = 20.19⁰

From the diagram attached;

  • assuming we have an imaginary position at the base of Ollie Standing point called point F when Ollie first activates the sensor;          

Then, we can say:

∠CBD = ∠GBF

∠GBF = (CAB + ACB)      

(because the exterior angles of a Δ is the sum of the two interior angles.

∠GBF = 15° + 20.19°

∠GBF = 35.19°

Using the trigonometric function for the tangent of an angle.

\mathbf{Tan \theta = \dfrac{GF}{BF}}

\mathbf{Tan \ 35.19  = \dfrac{1.8 \ m }{BF}}

\mathbf{BF  = \dfrac{1.8 \ m }{Tan \ 35.19}}

\mathbf{BF  = \dfrac{1.8 \ m }{0.7052}}

BF = 2.55 m

Finally, the distance of Ollie║FE║ from the entrance of his bouse is:

= 5 - 2.55 m

= 2.45 m

Therefore, we can conclude that Ollie is at a distance of 2.45 m from the entrance to his house when he first activates the sensor.

Learn more about exterior angles here:

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