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mihalych1998 [28]

3 years ago

13

A water sprinkler sends water out in a circular pattern. How many feet away from the sprinkler can it spread water if the area f

ormed by the watering pattern is 1 comma 017.36 square feet? Use 3.14 for pi.

2 answers:

nlexa [21]3 years ago

7 0

Answer:

The sprinkler can spread water 18 feet away.

Step-by-step explanation:

We are given the following in the question:

Area formed by watering pattern = 1,017.36 square feet

We have to find the how far the sprinkler spread the water.

The sprinkler covers a circular area. We need to find the radius of this circular area to find the how far the sprinkler spread the water.

Area of circle = $A = \pi r^2$

Putting values, we get:

$1017.36 = 3.14 r^2\\\\r^2 = \dfrac{1017.36}{3.14}\\\\r^2 = 324\\r = \sqrt{324} = 18\text{ feet}$

Therefore, the sprinkler can spread water 18 feet away.

myrzilka [38]3 years ago

6 0

Answer:

The sprinkler can spread water 18 feet away.

Step-by-step explanation:

We are given the following in the question:

Area formed by watering pattern = 1,017.36 square feet

We have to find the how far the sprinkler spread the water.

The sprinkler covers a circular area. We need to find the radius of this circular area to find the how far the sprinkler spread the water.

Area of circle =

$A = \pi r^2$

where r is the radius of the circle.

Putting values, we get,

$1017.36 = 3.14 r^2\\\\r^2 = \dfrac{1017.36}{3.14}\\\\r^2 = 324\\r = \sqrt{324} = 18\text{ feet}$

Thus, the sprinkler can spread water 18 feet away.

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3 years ago

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Answer:

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Step-by-step explanation:

Let L be the length of the ladder,

Given,

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$=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)$

$=\frac{1}{2\sqrt{x^2 + y^2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})$

$=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Here,

$\frac{dy}{dt}=-6\text{ ft per sec}$

Also, $\frac{dL}{dt} = 0$ ( Ladder length = constant ),

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8 0

3 years ago

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Answer:

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Step-by-step explanation:

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7 0

2 years ago

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Answer the Blanks:

vfiekz [6]

Answer:

1. 90°

2. 90°

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4. 45°

5. 45°

Step-by-step explanation:

<u>Given:</u> ΔABC,

CD⊥AB,

m∠A=50°,

m∠ACB=85°

<u>Solution:</u>

1. ∠ADC is a right ange, because CD⊥AB, so

$m\angle ADC=90^{\circ}$

2. ∠CDB is a right ange, because CD⊥AB, so

$m\angle CDB=90^{\circ}$

3. Consider triangle ACD. The sum of the measures of all interior angles is always 180°, so

$m\angle A+m\angle ADC+m\angle ACD=180^{\circ}\\ \\50^{\circ}+90^{\circ}+m\angle ACD=180^{\circ}\\ \\m\angle ACD=180^{\circ}-50^{\circ}-90^{\circ}=40^{\circ}$

4. By Angle Addition Postulate,

$m\angle ACD+m\angle BCD=m\angle ACB\\ \\40^{\circ}+m\angle BCD=85^{\circ}\\ \\m\angle BCD=85^{\circ}-40^{\circ}=45^{\circ}$

5. Consider triangle ABC. The sum of the measures of all interior angles is always 180°, so

$m\angle A+m\angle ACB+m\angle B=180^{\circ}\\ \\50^{\circ}+85^{\circ}+m\angle B=180^{\circ}\\ \\m\angle B=180^{\circ}-50^{\circ}-85^{\circ}=45^{\circ}$

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3 years ago

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