50 ÷ 10 = 5
You are running 5 meters per second.
An envelope is rectangular in shape.
Given the width = 15cm, and diagonal = 17cm
Let h represent the tall length of the envelope
Applying Pythagoras theorem, we have
![\begin{gathered} 17^2=15^2+h^2 \\ 289=225+h^2 \\ h^2=289-225 \\ h^2=64 \\ h=\sqrt[]{64} \\ h=8\operatorname{cm} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%2017%5E2%3D15%5E2%2Bh%5E2%20%5C%5C%20289%3D225%2Bh%5E2%20%5C%5C%20h%5E2%3D289-225%20%5C%5C%20h%5E2%3D64%20%5C%5C%20h%3D%5Csqrt%5B%5D%7B64%7D%20%5C%5C%20h%3D8%5Coperatorname%7Bcm%7D%20%5Cend%7Bgathered%7D)
The envelope is 8cm tall
Answer:
To find the area of a regular hexagon, or any regular polygon, we use the formula that says Area = one-half the product of the apothem and perimeter. As shown below, this means that we must find the perimeter (distance all the way around the hexagon) and the measure of the apothem using right triangles and trigonometry. Area of a Hexagon
Step-by-step explanation:
