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7nadin3 [17]
2 years ago
13

The scale model of the car is 1:18 the length of the model is 9 1/2 inches how many feet long is the actual

Mathematics
2 answers:
galina1969 [7]2 years ago
6 0

Answer: The length of the car is 14 and 1/4 (or 14.25 ) feets long.

Step-by-step explanation:

The scale is 1:18.

The lenght of the car is 9 and 1/2 inches.

Then, for the ratio relation, we have that each inch in the model car we have that corresponds to 18 inches in the car, so the length of the car is:

L = (9 +1/2)*18 in = 9.5*18 in = 171 inches.

Now, remember that 1ft = 12 inches, so if we want to write this in feet we must divide this by 12.

L = (171/12) ft = 14.25 ft = (14 + 1/4) ft

Y_Kistochka [10]2 years ago
5 0

Answer:

14.25 ft

Step-by-step explanation:

1 inch of the model represents 18 inches of the real car. The car's dimensions are 18 times larger than the model's dimensions. Multiply the model dimension by 18 to find the corresponding real dimension.

9 1/2 * 18 = 9.5 * 18 = 171

The actual car is 171 inches long.

Now we need to convert 171 inches to ft.

1 ft = 12 in.

Since we want to eliminate inches and end up with feet, we divide both sides by 12 in.

(1 ft)/(12 in.) = 1

The conversion factor is (1 ft)/(12 in.) to go from inches to feet.

171 in. * (1 ft)/(12 in.) = 14.25 ft

Answer: 14.25 ft

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kotegsom [21]

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x=-4.36

     

Step-by-step explanation:

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2 years ago
Find f(-5) if f( x ) = | x + 1|.
Alekssandra [29.7K]

Answer:

Step-by-step explanation:

f(x)=x+1

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2 years ago
A horseback riding trail is 1 mi 40.8 ft long. How long is the trail in yards, feet, and inches?
ch4aika [34]

Answer:  1773.6 yards   5320.8 feet  64324.8 inches

Step-by-step explanation:  

1 mile in yards = 1760                                                                                         40.8 feet in yards = 13.6      1760 + 13.6 =<u> 1773.6 yards </u>

1 mile in feet = 5280 + 40.8 = <u> 5320.8 feet </u>

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40.8 feet in inches = 964.8

63360 + 964.8 = <u>64324.8 inches </u>

<u>hope this helps :) </u>



6 0
3 years ago
Determine the formula for the nth term of the sequence:<br>-2,1,7,25,79,...​
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A plausible guess might be that the sequence is formed by a degree-4* polynomial,

x_n = a n^4 + b n^3 + c n^2 + d n + e

From the given known values of the sequence, we have

\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}

Solving the system yields coefficients

a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4

so that the n-th term in the sequence might be

\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}

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\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of \{x_n\} eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of \{x_n\} by \Delta^{k}\{x_n\}. Then

• 1st-order differences:

\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}

• 2nd-order differences:

\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}

• 3rd-order differences:

\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}

• 4th-order differences:

\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}

From here I made the assumption that \Delta^4\{x_n\} is the constant sequence {15, 15, 15, …}. This implies \Delta^3\{x_n\} forms an arithmetic/linear sequence, which implies \Delta^2\{x_n\} forms a quadratic sequence, and so on up \{x_n\} forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.

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