Q = 3s - 9/6
q = 3(s - 3)/6
q = s - 3/2
6q + 9 = 3s
6q + 9/3 = s
3(2q + 3)/3 = s
2q + 3 = s
s = 2q + 3
Answer:
144=12², so 144^3/2=(12²)^(3/2)=12³=1728.
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To find the area of his exclusion zone you would need to understand that a triangle with dimensions of 3, 4, and 5 represent a right triangle.
This means the exclusion zone would be applied to the base and the height of the triangular space.
You would add 2 km to the 3 km, and 2 km to the 4 km to create a new height of 5 km and a new base of 6 km.
Please see the attached picture to understand this.
You will find the area of the total space created by the new triangle and subtract the space represented by the original triangle to find the area of the exclusion zone.
(1/2 x 6 x 5) - (1/2 x 4 x 3)
15 km² -6 km² equals 9 km².
The exclusion space is 9 km².
Answer:
Step-by-step explanation:
The ratio of corresponding sides DN and KI is 12 : 4 = 3 : 1. The same ratio applies to altitudes DQ and KO. Since the difference between these altitudes is 6 and the difference between their ratio units is 3-1 = 2, each ratio unit must stand for 6/2 = 3 units of linear measure. That is, ...
DQ = (3 units)·3 = 9 units
KO = (3 units)·1 = 3 units
Then the base lengths QN and OI can be found from the Pythagorean theorem:
KI² = KO² +OI²
4² = 3² +OI²
OI = √(16 -9)
OI = √7
QN = 3·OI = 3√7