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love history [14]
3 years ago
10

if the diameter of the front wheels on the locomotive is 4 ft what is the diameter of the front wheels on the model. Express the

answer in inches?
Mathematics
1 answer:
Leto [7]3 years ago
6 0

Question is Incomplete, Complete question is given below.

Consider a model of a train locomotive that has a similarity transformation map of x arrow right 48x from the model to the actual locomotive when answering the question.

If the diameter of the front wheels on the locomotive is 4 feet, what is the diameter of the front wheels on the model? Express the answer in inches.

Answer:

The diameter of front wheel on the model is 1 inch.

Step-by-step explanation:

Given:

Let diameter of the front wheel on the model be 'x'.

Actual diameter of front wheel = 4 feet

Now given that ratio model to actual = \frac{x}{48x}= \frac{1}{48}

Now we can say that  ratio model to actual is equal to ratio of model diameter of front wheel to actual diameter of front wheel.

framing in equation form we get;

\frac{x}{4}=\frac{1}{48}\\\\x=\frac{4}{48} = \frac{1}{12} \ ft.

Now we know that;

\frac{1}{12}\ ft = 1 \ in.

Hence The diameter of front wheel on the model is 1 inch.

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(√3/2 a, a/2,a) are the sides of the triangle.


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How much money has to be invested at 5.9% interest compounded continuously to have $15,000 after 12 years?
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The amount of $7389.43 has to be invested at 5.9% interested continuously to have $15,000 after 12 years.

Step-by-step explanation:

The given is,

                Future value, F  = $15,000

                           Interest, i = 5.9%

              ( compounded continuously )

                            Period, t = 12 years

Step:1

           Formula to calculate the present with compounded continuously,

                                       F=Pe^{(i)(t)}...............(1)

           Substitute the values in equation (1) to find the P value,

                                  15000=Pe^{(0.059)(12)}          ( ∵ i = \frac{5.9}{100}=0.059 )

                                  15000=Pe^{0.708}

                                  15000=P(2.0299)             ( ∵ e^{o.708} =2.0299 )

            We change the P (Present value) into the left side,

                                        P=\frac{15000}{2.0299}

                                            =7389.427

                                            ≅ 7389.43

                                         P = $ 7389.43

Result:

           The amount of $7389.43 has to be invested at 5.9% interested continuously to have $15,000 after 12 years.  

                       

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