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elena-s [515]
3 years ago
8

The cost in dollars, y, to participate x days in a bike tour is a linear function. Lin's bike tour is described by the equation

y = 5x + 50. The cost of Max's bike tour is $120 for a 14-day bike tour. If both Lin and Max take 14-day tours, whose bike tour costs less? Lin Max They cost the same amount
Mathematics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

They cost the same amount

Step-by-step explanation:

Given

Lin: y = 5x +50 where x = 14

Max: \$120 for 14 days

Required

Which is less expensive?

Max charges has been given from the question as; $120 for 14 days.

To determine Lin's, we simply substitute 14 for x in y = 5x +50

y = 5  * 14 + 50

y = 70 + 50

y =120

The above result of y shows that Max charges for 14 days is also $120.

<em>Hence, they (Lin and Max) cost the same because they both equal to $120 for 14 days</em>

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A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and
nasty-shy [4]

Answer:

correct option is C)  2.8

Step-by-step explanation:

given data

string vibrates form =  8 loops

in water loop formed =  10 loops

solution

we consider  mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = \rho

case (1)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

so here

l = \frac{8 \lambda _1}{2}      ..............1

l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}

and we know velocity is express as

velocity = frequency × wavelength   .....................2

\sqrt{\frac{Tension}{mass\ per\ unit \length }}   =   f × \lambda_1

here tension = mg

so

\sqrt{\frac{mg}{\mu}}   =   f × \lambda_1     ..........................3

and

case (2)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

l = \frac{10 \lambda _1}{2}      ..............4

l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}

when block is immersed

equilibrium  eq will be

Tenion + force of buoyancy = mg

T + v × \rho × g = mg

and

T = v × \rho - v × \rho × g    

from equation 2

f × \lambda_2 = f  × \frac{1}{5}  

\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}     .......................5

now we divide eq 5 by the eq 3

\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}

solve irt we get

1 - \frac{\rho _{stone}}{\rho _{water}}  = \frac{16}{25}

so

relative density \frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}

relative density = 2.78 ≈ 2.8

so correct option is C)  2.8

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