150 pounds to 135 pounds identify
1 answer:
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Let
x-------> the first number
y-------> the second number
we know that
equation 1
equation 2
substitute the equation 1 in equation 2
![[-5-y]*y=-90 \\ -5y- y^{2} =-90 \\ y^{2} +5y-90=0](https://tex.z-dn.net/?f=%5B-5-y%5D%2Ay%3D-90%20%5C%5C%20-5y-%20y%5E%7B2%7D%20%3D-90%20%5C%5C%20y%5E%7B2%7D%20%2B5y-90%3D0)
using a graphical tool to resolve the second order equation
see the attached figure
the numbers are
-12.311
7.311
x-------> the first number
y-------> the second number
we know that
substitute the equation 1 in equation 2
using a graphical tool to resolve the second order equation
see the attached figure
the numbers are
-12.311
7.311
The correct answer is the flagpole is <span>33 feet high</span>.
Explanation:
Please refer to the attached picture.
We know:
CD = 40 feet
AC = 5 feet
∠BDC = α = 35°
Using trigonometry, we know that the definition of the tangent of an angle is the ratio between the opposite side and the adjacent side, therefore:
tan α = BC / CD
Solving for BC:
BC = CD · <span>tan α
= 40 </span>· tan (35)
= 28 feet
In order to find the height of the flagpole, we need to add the distance of the clinometer from the ground:
AB = BC + AC
= 28 + 5
= 33
Hence, the flagpole is 33 feet high.
Explanation:
Please refer to the attached picture.
We know:
CD = 40 feet
AC = 5 feet
∠BDC = α = 35°
Using trigonometry, we know that the definition of the tangent of an angle is the ratio between the opposite side and the adjacent side, therefore:
tan α = BC / CD
Solving for BC:
BC = CD · <span>tan α
= 40 </span>· tan (35)
= 28 feet
In order to find the height of the flagpole, we need to add the distance of the clinometer from the ground:
AB = BC + AC
= 28 + 5
= 33
Hence, the flagpole is 33 feet high.