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4 years ago

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A model rocket is 3 inches long. the real rocket is 120 ft long. what is the unit rate that describes the relationship of the re

al rocket to the model rocket?
A. 360 ft/in
B. 360 in/ft
C. 40 ft/in
D. 40 in/ft

2 answers:

V125BC [204]4 years ago

6 0

Answer: Option C

40 ft/in

Step-by-step explanation:

We know that:

model rocket = 3 inches long

real rocket = 120 ft long

To find the unit rate that describes the relationship of the real rocket to the model rocket, you must divide the length of the real rocket by the length of the model rocket:

$r =\frac{120\ ft}{3\ in}$

$r =40\frac{ft}{in}$

This means that every inch in length of the model rocket equals 40 feet of real rocket length

The answer is the option C

lora16 [44]4 years ago

4 0

Answer:

40 ft/in

Step-by-step explanation:

There is 40 feet for every inch of the model

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3 years ago

If the original square had a side length of

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Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

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State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

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The trinomial is

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therefore

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