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sweet-ann [11.9K]
2 years ago
5

Q. You are working on an air conditioning system. A roll of cylindrical copper tubing has an outside diameter of 7/8 inch and an

inside diameter of 3/4 inch. How much refrigerant can 12 feet of the tubing hold?
Mathematics
1 answer:
diamong [38]2 years ago
5 0

The refrigerant can hold 1.92 ft³

Step-by-step explanation:

The formula to apply is that of volume

v=π *r²*h where v is volume, r is radius of cylindrical object and h is height   of tubing

Finding volume of using outside  diameter

v=π*7/16*7/16*12

v=49/64 *3/1 *3.14

v=7.22 ft³

Volume using inside diameter

v=π*r²*h

v=3.14*3/8*3/8*12

v=5.30 ft³

Volume the refrigerant  can hold is;

7.22-5.30 = 1.92 ft³

Learn More

Volume of a cylindrical object:brainly.com/question/2665971

Keywords :conditioning,system,roll,copper tubing,refrigerant

#LearnwithBrainly

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mojhsa [17]
To determine the cost of each item, we need to set up equations. From the problem statement, we have three unknowns so we need three equations. We set up equations as follows:

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x + y + 3z = 15

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3x + y + 2z = 22

Solving for x, y and z, we will have:

x = $ 5
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An isosceles triangle ABC with the base BC is inscribed in a circle. Find the measure of angles of it, if measure of arc BC = 10
Usimov [2.4K]

Answer:

The measures of the angles of the triangle are 51° , 64.5° , 64.5°

OR

The measures of the angles of the triangle are 129° , 25.5° , 25.5°

Step-by-step explanation:

* Lets explain the meaning of the inscribed triangle in a circle

- If a triangle inscribed in a circle, then the vertices of the triangle lie

 on the circumference of the circle and each vertex is an inscribed

 angle in the circle subtended by the opposite arc

- Fact in the circle the measure of the inscribed angle is 1/2 the

 measure of its subtended arc

* Now lets solve the problem

- Δ ABC is an isosceles with the base BC

∴ AB = AC

∴ m∠B = m∠C

- Δ ABC is inscribed in a circle

∴ ∠A is inscribed angle subtended by arc BC (minor or major)

# The measure of the minor arc is less than 180° and the measure of

  the major arc is greater then 180° and the sum of the two arcs

  equals the measure of the circle which is 360°

∴ ∠B subtended by arc AC

∴ ∠C subtended by arc AB

∵ The measure of the arc BC = 102°

- There is two cases in this question

(1) If the angle A subtended by the minor arc BC

(2) If the angle A subtended by the major arc BC

- Lets solve case (1)

∵ ∠A is an inscribed angle subtended by the minor arc BC

∴ m∠A = 1/2 the measure of the arc BC

∵ The measure of the arc BC is 102°

∴ m∠A = 1/2 × 102 = 51°

∵ The sum of the measures of the interior angles of a triangle is 180°

∴ m∠A + m∠B + m∠C = 180°

∴ 51 + m∠B + m∠C = 180° ⇒ subtract 51 from both sides

∴ m∠B + m∠C = 129°

∵ m∠B = m∠C ⇒ isosceles Δ

∴ m∠B = m∠C = 129/2 = 64.5°

* The measures of the angles of the triangle are 51° , 64.5° , 64.5°

- Lets solve case (2)

∵ ∠A is an inscribed angle subtended by the major arc BC

∴ m∠A = 1/2 the measure of the arc BC

∵ The measure of the minor arc BC is 102°

∵ The measure of the circle is 360°

∴ The measure of the major arc = 360 - 102 = 258°

∴ m∠A = 1/2 × 258 = 129°

∵ The sum of the measures of the interior angles of a triangle is 180°

∴ m∠A + m∠B + m∠C = 180°

∴ 129 + m∠B + m∠C = 180° ⇒ subtract 129 from both sides

∴ m∠B + m∠C = 51°

∵ m∠B = m∠C ⇒ isosceles Δ

∴ m∠B = m∠C = 51/2 = 25.5°

* The measures of the angles of the triangle are 129° , 25.5° , 25.5°

4 0
2 years ago
Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
soldier1979 [14.2K]

x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0
2 years ago
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