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2 years ago

A standard fire hose is 50 feet long. What is the length of the hose in yards and feet?

Mathematics

2 answers:

Tpy6a [65]2 years ago

3 0

50 feet = 16.67 yards

Elodia [21]2 years ago

3 0

16.7 rounded if you round the number to a tens place I think if I’m not mistaken

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Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

2 years ago

Help please someone help

MariettaO [177]

Answer:

KJZV JCH BSV

Step-by-step explanation:

CBZ,NSFVN bvVVVVV VVVVV VSFNV.Z

3 0

2 years ago

A sphere is inscribed in a cylinder. The surface area of the sphere and the lateral area of the cylinder are equal. true or fals

Volgvan

It is true that the surface area of the sphere and the lateral area of the cylinder are equal. The answer is True.

8 0

3 years ago

Read 2 more answers

1. Name the figure (It is not a cube; it is not a cylinder; what is it?)

Arlecino [84]

1. It is a triangular prism

6 0

2 years ago

If ΔLMN ≅ ΔOPR, m∠L = (x^2 - x)°, m∠P = 16°, m∠N = (4x+160)°. Find m∠R.

Marizza181 [45]

Answer:

mZR = (4x + 160)°

Step-by-step explanation:

cause, mZR = mZN

;-))

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2 years ago