Answer:
22m
Step-by-step explanation:
Let height of flagpole=h
AB==17 m
(1 degree= 60 minute)
![\angle B=25^{\circ}43'=25+\frac{43}{60}=25.72^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20B%3D25%5E%7B%5Ccirc%7D43%27%3D25%2B%5Cfrac%7B43%7D%7B60%7D%3D25.72%5E%7B%5Ccirc%7D)
We have to find the approximate height of the flagpole.
In triangle CDA,
![\frac{CD}{DA}=tan\theta=\frac{Perpendicular\;side}{Base}](https://tex.z-dn.net/?f=%5Cfrac%7BCD%7D%7BDA%7D%3Dtan%5Ctheta%3D%5Cfrac%7BPerpendicular%5C%3Bside%7D%7BBase%7D)
![h=DA(0.759)](https://tex.z-dn.net/?f=h%3DDA%280.759%29)
In triangle CDB,
![tan 25.72^{\circ}=\frac{CD}{DB}](https://tex.z-dn.net/?f=tan%2025.72%5E%7B%5Ccirc%7D%3D%5Cfrac%7BCD%7D%7BDB%7D)
![0.482=\frac{h}{DA+17}](https://tex.z-dn.net/?f=0.482%3D%5Cfrac%7Bh%7D%7BDA%2B17%7D)
![0.482DA+8.194=h](https://tex.z-dn.net/?f=0.482DA%2B8.194%3Dh)
Substitute the value
![0.482DA+8.194=0.759DA](https://tex.z-dn.net/?f=0.482DA%2B8.194%3D0.759DA)
![8.194=0.759DA-0.482DA](https://tex.z-dn.net/?f=8.194%3D0.759DA-0.482DA)
![8.194=0.277DA](https://tex.z-dn.net/?f=8.194%3D0.277DA)
![DA=\frac{8.194}{0.277}=29.58](https://tex.z-dn.net/?f=DA%3D%5Cfrac%7B8.194%7D%7B0.277%7D%3D29.58)
Substitute the value
![h=29.58 \times 0.759=22.45 m\approx 22m](https://tex.z-dn.net/?f=h%3D29.58%20%5Ctimes%200.759%3D22.45%20m%5Capprox%2022m)
Hence, the height of the flagpole=22 m
Answer:
see below
Step-by-step explanation:
The functions for these tables are, respectively, ...
f(x) = 5 - x
f(x) = 3x
f(x) = (x -1)² . . . . . a quadratic relation
f(x) = 4x² . . . . . . a quadratic relation
We assume the intention is that the terminology "quadratic variation" mean "proportional to the square of x". In that case, only the last function has such variation.
Answer:
it's a
Step-by-step explanation:
4 would be the whole number. 0.8 would be 4/5 so the mixed number is 4 and 4/5.