Answer:
I'm going to use the Pythagoras theorem
1. hypotenus²= base²+height²
h²=80²+18²
h²=6400+324
√h²=√6724
hypotenuse= 82
2. 53²=45²+h²
2809=2025+h²
2809-2025=h²
784=h²
√784=√h²
h=28
3. 40²=b²+24²
1600=b²+576
1600-576=b²
√1024=√b²
b= 32
The farthest distance of the turtle can be solved with the following equations:
x = 112 + 4
x = 112 - 4
By solving the equations, we conclude that t<span>he turtle can be found either in the 116th block or the 108th block.</span>
Answer:
4n-13
Step-by-step explanation:
Terms= -13 (constant)
Coefficient= 3, -5, 6
Like term= 3n, -5n, 6n
3n-13-5n+6n
3n+6n-5n-13
9n-5n-13
4n-13
Hope this helps ;) ❤❤❤
Answer:
Altitude of the jet is 10.40km
Step-by-step explanation:
The angle of elevation = 8°
Horizontal distance (adjacent to the angle) = 74km
Altitude of the plane (opposite) = ?
Kindly check attached document for better illustration.
To solve this question, we have to use tangent rule since we have adjacent and angle, but no opposite.
Tanθ = opposite / adjacent
θ = 8°
Adjacent = 74km
Opposite = ?
Tan 8 = x / 74
Opposite = 74 * tan8
Opposite = 10.40
The altitude of the jet is 10.40km