Answer:
The actual height of the object is 200.2 cm
Step-by-step explanation:
The parameters given are;
The actual height = 154 cm.
The height in the picture = 10 cm.
The height of the object = 13 cm
The scale of the picture is 154 is to 10 or 154:10 which is equivalent to 77:5
Therefore the scale of the actual object height to the image height = 77:5
Where the image of the object = 13 cm
Let the height of the actual object = X, then we then have;
![\dfrac{77}{5} = \dfrac{X}{13}](https://tex.z-dn.net/?f=%5Cdfrac%7B77%7D%7B5%7D%20%3D%20%5Cdfrac%7BX%7D%7B13%7D)
![X = 13 \times \dfrac{77}{5} = \dfrac{1001}{5} = 200\frac{1}{5} \ cm = 200.2 \ cm](https://tex.z-dn.net/?f=X%20%3D%2013%20%5Ctimes%20%5Cdfrac%7B77%7D%7B5%7D%20%3D%20%5Cdfrac%7B1001%7D%7B5%7D%20%3D%20200%5Cfrac%7B1%7D%7B5%7D%20%20%5C%20cm%20%3D%20200.2%20%5C%20cm)
The actual height of the object = 200.2 cm.
Answer:
→ ![4\sqrt{3}](https://tex.z-dn.net/?f=%204%5Csqrt%7B3%7D%20)
Step-by-step explanation:
Given:
![\sqrt{48}](https://tex.z-dn.net/?f=%20%5Csqrt%7B48%7D%20)
Get, prime factorisation of 48
So, factors are 2×2×2×2×3
Make pairs → ![2 \: \times \: 2 \: \sqrt{3}](https://tex.z-dn.net/?f=2%20%5C%3A%20%20%5Ctimes%20%20%5C%3A%202%20%5C%3A%20%5Csqrt%7B3%7D%20)
» ![4 \sqrt{3}](https://tex.z-dn.net/?f=4%20%5Csqrt%7B3%7D%20)
or 6.9
Not 100% sure if this is how you do it but I tried.
Answers:
10mi
approx. 42.98ft.
60yd
approx. 73.31ft.
approx. 0.90ft.
approx. 12.37ft.
<span>The area of a blade with rectangular shape = 1/2 * b * h where b is the base and h, the height.
Here b = 7 and height = 4
So the area of the blade = 1/2 * 7 * 4 = 14ft^2
If the are four blades the total area is the summation of individual blade.
So area = 14 + 14 + 14 + 14 = 56ft^2. Hence none of the option satisfies the answer.</span>
It would still be 75, because translations preserve angle measures