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3 years ago

the volume of a cylinder is 44cm3. find the volume of another cylinder of the same height and double the base radius

Mathematics

2 answers:

miskamm [114]3 years ago

5 0

So what you need to do is form another cylinder that has a volume of 44cm3 and then double the base size by the radius.

Hoochie [10]3 years ago

5 0

Answer:

$Volume \ of \ other\ cylinder = 176 \ cm^3$

Step-by-step explanation:

Let the volume of cylinder Vₐ = 44cm³

Let radius of cylinder " a " be = rₐ

Let height of cylinder " b" be = hₐ

$Volume_a = \pi r_a^2 h_a\\\\44 = \pi r_a^2 h_a$

Given cylinder " b ", Radius is twice cylinder " a " , that is $r_b = 2 r_a$

Also Height of cylinder " b " is same as cylinder " a " , that is $h_b = h_a$

$Volume_b = \pi r_b^2 h_b$

$= \pi (2r_a)^2 h_a\\\\=4 \times \pi r_a^2 h_a\\\\= 4 \times 44\\\\= 176 \ cm^3$

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3 years ago

The volume of a rectangular box with a square base remains constant at 500 cm3 as the area of the base increases at a rate of 6

Andre45 [30]

Answer:

the rate at which the height of the box is decreasing is -0.0593 cm/s

Step-by-step explanation:

Given the data in the question;

Constant Volume of a rectangular box with a square base = 500 cm³

area of the base increases at a rate of 6 cm²/sec

so change in the area of the base with respect to time dA/dt = 6 cm²/sec

each side of the base is 15 cm long

so Area of the base = 15 cm × 15 cm = 225 cm²

the rate at which the height of the box is decreasing = ?

Now,

V = Ah

dv/dt = 0 ⇒ Adh/dt + hdA/DT = 0

⇒ dh/dt = -hdA/dt / A

we substitute

dh/dt = [ -( 500 / 225 ) × 6 ] / 225

dh/dt = [ -(2.22222 × 6) ] / 225

dh/dt = [ -13.3333 ] / 225

dh/dt = -0.0593 cm/s

Therefore, the rate at which the height of the box is decreasing is -0.0593 cm/s

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3 years ago

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dimulka [17.4K]

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2 years ago

Find the following quotients, and write the quotient in standard form. (a)- x^2-9 ÷ x-3 (b)- x^3-27 ÷ x-3 (c) x^4- 81 ÷ x-3

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Answer:

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Step-by-step explanation:

We have given that

(a) $\frac{x^2-9}{x-3}$

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$a^2-b^2=(a+b)(a-b)$

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(b) $\frac{x^3-27}{x-3}$

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So $\frac{x^3-27}{x-3}=\frac{x^3-3^3}{x-3}=\frac{(x-3)(x^2+3x+9)}{x-3}=x^2+3x+9$

We know the algebraic identity

$a^2-b^2=(a+b)(a-b)$

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