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Step-by-step explanation:hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
Height of another tree that cast a shadow which is 20ft long is 5 feet approximately
<h3><u>Solution:</u></h3>
Given that tree with a height of 4 ft casts a shadow 15ft long on the ground
Another tree that cast a shadow which is 20ft long
<em><u>To find: height of another tree</u></em>
We can solve this by setting up a ratio comparing the height of the tree to the height of the another tree and shadow of the tree to the shadow of the another tree
![\frac{\text {height of tree}}{\text {length of shadow}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bheight%20of%20tree%7D%7D%7B%5Ctext%20%7Blength%20of%20shadow%7D%7D)
Let us assume,
Height of tree = ![H_t = 4 feet](https://tex.z-dn.net/?f=H_t%20%3D%204%20feet)
Length of shadow of tree = ![L_t = 15 feet](https://tex.z-dn.net/?f=L_t%20%3D%2015%20feet)
Height of another tree = ![H_a](https://tex.z-dn.net/?f=H_a)
Length of shadow of another tree = ![L_a = 20 feet](https://tex.z-dn.net/?f=L_a%20%3D%2020%20feet)
Set up a proportion comparing the height of each object to the length of the shadow,
![\frac{\text {height of tree}}{\text {length of shadow of tree}}=\frac{\text { height of another tree }}{\text { length of shadow of another tree }}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bheight%20of%20tree%7D%7D%7B%5Ctext%20%7Blength%20of%20shadow%20of%20tree%7D%7D%3D%5Cfrac%7B%5Ctext%20%7B%20height%20of%20another%20tree%20%7D%7D%7B%5Ctext%20%7B%20length%20of%20shadow%20of%20another%20tree%20%7D%7D)
![\frac{H_{t}}{L_{t}}=\frac{H_{a}}{L_{a}}](https://tex.z-dn.net/?f=%5Cfrac%7BH_%7Bt%7D%7D%7BL_%7Bt%7D%7D%3D%5Cfrac%7BH_%7Ba%7D%7D%7BL_%7Ba%7D%7D)
Substituting the values we get,
![\frac{4}{15} = \frac{H_a}{20}\\\\H_a = \frac{4}{15} \times 20\\\\H_a = 5.33](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B15%7D%20%3D%20%5Cfrac%7BH_a%7D%7B20%7D%5C%5C%5C%5CH_a%20%3D%20%5Cfrac%7B4%7D%7B15%7D%20%5Ctimes%2020%5C%5C%5C%5CH_a%20%3D%205.33)
So the height of another tree is 5 feet approximately
The normal body temperature : to = 37° C
An inequality is:
| t - to | ≤ 3
| t - 37 | ≤ 3
t - 37 ≤ 3 AND ( - t - 37 ) ≤ 3
t ≤ 3 +
37
- t + 37 ≤ 3
t ≤
40
- t ≤ 3 - 37
- t ≤ - 34 / *( - 1 )
t ≥ 34
Answer t <span>∈ [</span><span>
34</span><span>°C,</span><span>
40</span><span>° C ]</span>.
The frustum can be considered as consisting of a square pyramid, less the
cut volume.
- The mass of the stand, is approximately 18.60905 kg
Reasons:
The given parameter of the frustum are;
The density of the frustum, ρ = 85 g/cm³
Height of pyramid, h = 9 cm
Side length of base = 10 cm
Height of frustum
By proportional shapes, the side length of the top of the frustum can be found as follows;
![\dfrac{4 \, cm}{9 \, cm} = \dfrac{x}{10 \, cm}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%20%5C%2C%20cm%7D%7B9%20%5C%2C%20cm%7D%20%3D%20%5Cdfrac%7Bx%7D%7B10%20%5C%2C%20cm%7D)
![x = \dfrac{4 \, cm}{9 \, cm} \times 10 \, cm = 4\frac{4}{9} \, cm](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B4%20%5C%2C%20cm%7D%7B9%20%5C%2C%20cm%7D%20%5Ctimes%2010%20%5C%2C%20cm%20%3D%204%5Cfrac%7B4%7D%7B9%7D%20%5C%2C%20cm)
![V = \dfrac{h}{3} \times \left(B_1 + B_2 + \sqrt{B_1 \times B_2} \right)](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7Bh%7D%7B3%7D%20%5Ctimes%20%5Cleft%28B_1%20%2B%20B_2%20%2B%20%5Csqrt%7BB_1%20%5Ctimes%20B_2%7D%20%5Cright%29)
B₁ = 10², B₂ = ![\left(4\frac{4}{9} \right)^2](https://tex.z-dn.net/?f=%5Cleft%284%5Cfrac%7B4%7D%7B9%7D%20%5Cright%29%5E2)
Therefore;
![V = \dfrac{4}{3} \times \left(10^2 + \left(4\frac{4}{9} \right)^2 + \sqrt{10^2 \times \left(4\frac{4}{9} \right)^2} \right) \approx 218.93](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7B4%7D%7B3%7D%20%5Ctimes%20%5Cleft%2810%5E2%20%2B%20%5Cleft%284%5Cfrac%7B4%7D%7B9%7D%20%20%5Cright%29%5E2%20%20%2B%20%5Csqrt%7B10%5E2%20%5Ctimes%20%20%5Cleft%284%5Cfrac%7B4%7D%7B9%7D%20%20%5Cright%29%5E2%7D%20%5Cright%29%20%5Capprox%20218.93)
The volume of the stand, V ≈ 218.93 cm³
Mass = Volume × Density
∴ Mass of the stand, m = 218.93 cm³ × 85 g/cm³ = 18,609.05 grams = 18.60905 kg.
Learn more here:
brainly.com/question/24323975