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zhenek [66]
3 years ago
8

Which of the following options is a better purchase for a bicycle? Option 1: A cash sale for $88

Mathematics
2 answers:
Korvikt [17]3 years ago
6 0

Answer:

The answer is option 3

Step-by-step explanation:

stellarik [79]3 years ago
3 0

<u><em>Answer:</em></u>

Option 3

<u><em>Explanation:</em></u>

<u>To get the value of the cheapest bike, best option, we will calculate the total cost of each bike:</u>

<u>i- Option 1:</u>

Total cost = $88

<u>ii- Option 2:</u>

$5 down payment and $8 per week for 10 weeks

Total cost = 5 + 8(10) = 5 + 80 = $85

<u>iii- Option 3:</u>

$12 down payment and $5 per month for 12 months

Total cost = 12 + 5(12) = 12 + 60 = $72

<u>iv- Option 4:</u>

$20 down payment and $20 per month for 12 months

Total cost = 20 + 20(12) = 20 + 240 = $260

<u>Now, from the above calculations, we can conclude that:</u>

The best price would be that of option 3 ($72). It is the lowest price compared to other option.

Hope this helps :)

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The proof that ΔACB ≅ ΔECD is shown. Given: AE and DB bisect each other at C. Prove: ΔACB ≅ ΔECD What is the missing statement i
GarryVolchara [31]

Answer:

The missing statement is ∠ACB ≅ ∠ECD

Step-by-step explanation:

Given two lines segment AC and BD bisect each other at C.

We have to prove that ΔACB ≅ ΔECD

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Soap films and bubbles are colorful because the interference conditions depend on the angle of illumination (which we aren't cov
mylen [45]

Answer:

56.39 nm

Step-by-step explanation:

In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,

2t=(m+\frac{1}{2} ).\frac{\lambda}{n}

where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm

Here we take m=0.

Similarly for the constructive reflection at the blue end (400 nm)

t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm

Hence the thickness difference should be

t_1-t_2=131.58-75.19=56.39 \ nm

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