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2 years ago

Plz answer quickly!!!!!!!

Mathematics

1 answer:

ValentinkaMS [17]2 years ago

6 0

Answer:

(A). x ≥ $\frac{16}{5}$ and x ≤ - $\frac{3}{4}$

Step-by-step explanation:

5x - 4 ≥ 12 ⇔ 5x ≥ 16 ⇒ x ≥  $\frac{16}{5}$  

and

12x + 5 ≤ - 4 ⇔ 12x ≤ - 9 ⇔ 4x ≤ - 3 ⇒ x ≤ -  $\frac{3}{4}$

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If 300% of 0.18 is equivalent to 20% of b, then b is equivalent to what number?

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Answer:

2.7

Step-by-step explanation:

3*0.18=0.54

0.54/0.2=2.7

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2 years ago

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Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufa

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Complete Question

Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufactures such dewars of different sizes. Shown here are two dewars, both of the same height but of different diameters. How much aluminium will be required to cover Dewar A compared to that required for Dewar B? (Note: The base and the lid are NOT made up of aluminium.)

A) 1/3 times

B) 3 times

C) 9 times

D) The same amount of aluminum will be required.

Answer:

B) 3 times

Step-by-step explanation:

From the question, we are told that we are to find the amount of aluminum to cover Dewar A and B , we are also told that the base and the lid are not covered with aluminum.

From this information, we can determine that we are to find the curved or Lateral surface area of the cylinder because the base and lid are not included

The curved surface area of a cylinder is calculated as 2πrh

We told that both cylinders have the same height. Hence,

For Dewar A

We have Diameter of 30 cm, radius = Diameter ÷ 2 = 30cm ÷ 2 = 15cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 15 ×h

= 30πhcm²

For Dewar B

We have Diameter of 10 cm, radius = Diameter ÷ 2 = 10cm ÷ 2 = 5cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 5 ×h

= 10πhcm²

When we compare the curved surface Dewar A to the curved surface area of Dewar B

Dewar A : Dewar B

30πhcm² : 10πhcm²

3(10πhcm²) : 10πhcm²

From the above comparison, we can see that 3 times more aluminum is required to cover Dewar A compared to Dewar B.

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2 years ago

Two cross sections of a right hexagonal pyramid are obtained by cutting the pyramid with planes parallel to the hexagonal base.

Tanya [424]

Answer:

The larger cross section is 24 meters away from the apex.

Step-by-step explanation:

The cross section of a right hexagonal pyramid is a hexagon; therefore, let us first get some things clear about a hexagon.

The length of the side of the hexagon is equal to the radius of the circle that inscribes it.

The area is

$A=\frac{3\sqrt{3} }{2} r^2$

Where $r$ is the radius of the inscribing circle (or the length of side of the hexagon).

Now we are given the areas of the two cross sections of the right hexagonal pyramid: $A_1=216\:ft^2\: \:\:\:A_2=486\:ft^2$

From these areas we find the radius of the hexagons:

$r_1=\sqrt{\frac{2A_1}{3\sqrt{3} } } =\sqrt{\frac{2*216}{3\sqrt{3} } }=\boxed{9.12ft}$

$r_2=\sqrt{\frac{2A_2}{3\sqrt{3} } } =\sqrt{\frac{2*486}{3\sqrt{3} } }=\boxed{13.68ft}$

Now when we look at the right hexagonal pyramid from the sides ( as shown in the figure attached ), we see that $r_1$ $r_2$ form similar triangles with length $H$