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N76 [4]
3 years ago
7

Barry wants to make a drawing that is the size of the original. If a tree in the original drawing is 14 inches tall and 5 inches

wide, what will be the length and width of the tree in Barry's drawing?

Mathematics
1 answer:
Oksanka [162]3 years ago
5 0
14/4 = 7/2
5/4 cannot be simplified. 
The tree's parameters will be 3 1/2 inches tall and 5/4 inches wide.
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Answer:

C is the correct answer.

Step-by-step explanation:

The reason it is C is because pi/the symbol on top is irrational.

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Step-by-step explanation:

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Points E, F, and D are on circle C, and angle G measures 60°. The measure of arc EF equals the measure of arc FD.
galben [10]

Answer:

∠EFD ≅ ∠EGD ⇒ A

Arc ED ≅ arc FD ⇒ C

m arc FD = 120° ⇒ E

Step-by-step explanation:

Let us revise some facts

  1. Equal chords subtended equal arcs
  2. The measure of an inscribed angle is one-half the measure of the central angle which subtended by the same arc
  3. The measure of a central angle is equal to the measure of its subtended arc
  4. If one angle of an isosceles triangle measure 60° then the triangle is equilateral
  5. The sum of the measures of the interior angles of any quadrilateral is 360°

In the quadrilateral  CDGE

∵ m∠G = 60°

∵ m∠GDC = m∠GEC = 90°

- By using the 5th rule above

∴ m∠G + m∠GDC + m∠DCE + m∠GEC = 360°

∴ 60 + 90 + m∠DCE + 90 = 360

∴ 240 + m∠DCE = 360

- Subtract 240 from both sides

∵ m∠DCE = 120°

In circle C

∵ ∠DCE is a central angle subtended by arc DE

∵ ∠DFE is an inscribed angle subtended by arc DE

- By using the 2nd rule above

∴ m∠DFE = \frac{1}{2} m∠∠DCE

∵ m∠DCE = 120°

∴ m∠DFE = \frac{1}{2} (120)

∴ m∠DFE = 60°

- That means ∠EFD ≅ ∠EGD because their measure is 60°

∴ ∠EFD ≅ ∠EGD

In Δ EFD

∵ EF = FD

∵ m∠DFE = 60°

- By using the 4th rule above

∴ Δ EFD is an equilateral triangle

∴ ED = FD = FE

In circle C

∵ Side ED subtended by arc ED

∵ Side FD subtended by FD

∵ Side ED ≅ side FD ⇒ proved

- By using the 1st rule above

∴ Arc ED ≅ arc FD

∵ m∠ECD = 120°

∵ ∠ECD is a central angle subtended by arc ED

- By using the 3rd rule above

∴ m∠ECD = m arc ED

∴ m of arc ED = 120°

∵ Arc ED ≅ arc FD

∴ m arc ED = m arc FD

∴ m arc FD = 120°

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2 years ago
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Vlad [161]
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3 years ago
A cylinder shaped can needs to be constructed to hold 600 cubic centimeters of soup. The material for the sides of the can costs
PSYCHO15rus [73]

Answer:

the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm

Step-by-step explanation:

since the volume of a cylinder is

V= π*R²*L → L =V/ (π*R²)

the cost function is

Cost = cost of side material * side area  + cost of top and bottom material * top and bottom area

C = a* 2*π*R*L + b* 2*π*R²

replacing the value of L

C = a* 2*π*R* V/ (π*R²) + b* 2*π*R²  = a* 2*V/R + b* 2*π*R²

then the optimal radius for minimum cost can be found when the derivative of the cost with respect to the radius equals 0 , then

dC/dR = -2*a*V/R² + 4*π*b*R = 0

4*π*b*R = 2*a*V/R²

R³ = a*V/(2*π*b)

R=  ∛( a*V/(2*π*b))

replacing values

R=  ∛( a*V/(2*π*b)) = ∛(0.03$/cm² * 600 cm³ /(2*π* 0.05$/cm²) )= 3.85 cm

then

L =V/ (π*R²) = 600 cm³/(π*(3.85 cm)²) = 12.88 cm

therefore the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm

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