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3 years ago

5

A model airplane is built at a scale of 1 inch to 6 feet. If the model plane is 8 inches long, how many feet long is the actual

airplane?
Type your answer as a whole number in the box below.

_____ feet

2 answers:

AnnyKZ [126]3 years ago

7 0

Answer:

48 feet

Step-by-step explanation:

Let's create a proportion using the following setup.

inches/feet=inches/feet

We know the scale is 1 inch to 6 feet. We also know that the plane is 8 inches long, but we don't know how many feet long it is, so we can say <em>x </em>feet.

1 inch/6 feet= 8 inches/ x feet

1/6=8/x

Cross multiply. Multiply the numerator of the first fraction by the denominator of the second. Multiply the denominator of the first by the numerator of the second.

1*x=6*8

x=6*8

x=48

Add appropriate units, in this case: feet.

x= 48 feet

The actual airplane is 48 feet long.

noname [10]3 years ago

5 0

The airplane is 48 feet long in real dimensions.

To find out how long the airplane actually is, we can use the scale and input our amount of inches that it is measured by.

So, 1 inch equals to 6 feet.

Our model plane is 8 inches.

Now, we plug it in.

8 inches = 6(8)

8 inches = 48 feet

Therefore, the length of the real airplane is 48 feet.

48 feet

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Answer:

B. 21.64

Step-by-step explanation:

We have been given a triangle and we are asked to find the length of AC (b).

We will use law of cosines to find the length of side AC.

$c^{2}=a^{2}+b^{2}-2ab \text{ cos }\theta$

Upon substituting our given values in the formula we will get,

$(AC)^{2}=15^{2}+12^{2}-2\times 15\times 12 \text{ cos}(106)$

$(AC)^{2}=225+144-360\text{ cos}(106)$

$(AC)^{2}=369-360(-0.275637355817)$

$(AC)^{2}=369+99.22944809412$

$(AC)^{2}=468.22944809412$

Upon taking square root of both sides of our equation we will be get,

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blsea [12.9K]

Option A: $\sqrt{75}$

Option C: $\sqrt{15} \cdot \sqrt{5}$

Option F: $\sqrt{25} \cdot \sqrt{3}$

Solution:

Given expression is $5 \sqrt{3}$ .

Option A: $\sqrt{75}$

$\sqrt{75}=\sqrt{25\times3}$

$=\sqrt{5^2\times3}$

$=5\sqrt{3}$

Hence $\sqrt{75}$ is equivalent expression of $5 \sqrt{3}$ .

Option B: $\sqrt{45}$

$\sqrt{45}=\sqrt{9\times5}$

$=\sqrt{3^2\times5}$

$=3\sqrt{5}$

Hence $\sqrt{45}$ is not equivalent expression of $5 \sqrt{3}$ .

Option C: $\sqrt{15} \cdot \sqrt{5}$

$\sqrt{15} \cdot \sqrt{5}=\sqrt{15\times5}$

$=\sqrt{75}$

$=5\sqrt{3}$ (proved in option A)

Hence $\sqrt{15} \cdot \sqrt{5}$ is equivalent expression of $5 \sqrt{3}$ .

Option D: $\sqrt{3} \cdot \sqrt{5}$

$\sqrt{3} \cdot \sqrt{5}=\sqrt{3\times5}$

$=\sqrt{15}$

Hence $\sqrt{3} \cdot \sqrt{5}$ is not equivalent expression of $5 \sqrt{3}$ .

Option E: 75

75 is a whole number.

Hence 75 is not equivalent expression of $5 \sqrt{3}$ .

Option F: $\sqrt{25} \cdot \sqrt{3}$

$\sqrt{25} \cdot \sqrt{3}=\sqrt{25\times3}$

$=\sqrt{75}$

$=5\sqrt{3}$ (proved in option A)

Hence $\sqrt{25} \cdot \sqrt{3}$ is equivalent expression of $5 \sqrt{3}$ .

Therefore, $\sqrt{75},\ \ \sqrt{15} \cdot \sqrt{5}, \ \ \sqrt{25} \cdot \sqrt{3}$ are all equivalent expressions of $5 \sqrt{3}$ .

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