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3 years ago

Find the volume of the cone. Round to your nearest hundredth

Mathematics

1 answer:

Alika [10]3 years ago

8 0

volume of cone

= ⅓πr²h

= ⅓ × 3.14 × 9 × 9 × 12

= 3.14 × 3 × 108

= 3.14 × 324

= 1017.36 cm³

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What is the answer to this question?

Tanya [424]

Answer:

a^(-1/2) = 1/sqrt(a) = 3

so.

sqrt(a) = 1/3

a = 1/9

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3 years ago

Find the measure of angle x in the figure below<br>67 61<br>у<br>X<br>67

kondor19780726 [428]

Answer: y67

Step-by-step explanation:5n − 19 + n + 7 = 144 − 6n

6n − 12 = 144 − 6n

12n = 156

n = 13

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3 years ago

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Solve the following formula for x y = 7 m x − 8 b

katovenus [111]

Answer: It depends on which variable you're solving for.

Step-by-step explanation:

Solving for b: b=7/8mx + -1/8xy

Solving for m: m=xy+8b

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Solving for x: x=-8b

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-7m+y

Solving for y: y=7mx-8b

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7 0

2 years ago

Very important!! Solve the inequality 6r + 30 ≥ 12 or -r > 12 (the question is important for high school application c:)

Fofino [41]

First simplify the one inequality, 6r + 30 greater than or equal to 12 just divide by 6 then subtract 30 from both sides of the inequality. You should get r is greater than or equal to -3. For this one you need to reverse the inequality symbol then divide both sides by -1 to get a positive r, and you get r is less than -12. It's not A because there's two solutions, and it's not D for the same reason. It's not C because r isn't greater than -12. In conclusion it's B because it correctly represents the solutions that make the inequality true. I hope this helps you on your high school application :)

7 0

3 years ago

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$

Volume of the box at x= 1.59:

$V = \frac{1}{2} (5 – 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\$

Volume of the box at x= 0.53:

$V = \frac{1}{2} (5 – 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3$

The volume of the box is maximized when x = 0.53 cm

Therefore,

x = 0.53 cm

Maximum volume = 1.75 cm³

7 0

3 years ago