How are the graphs of the function f(x)= square root 16^x and g(x)=3 square root 64^x related
1 answer:
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I FOUND YOUR COMPLETE QUESTION IN OTHER SOURCES.
SEE ATTACHED IMAGE.
Using Heron's Formula we can find the area of the triangle.
A = root (s (s-a) * (s-b) * (s-c))
s = (a + b + c) / 2
Where,
s: semi-perimeter
a, b, c: sides of the triangle
Substituting values:
s = (15 + 16 + 20) / 2
s = 25.5
s = 26
The area will be:
A = root (26 * (26-15) * (26-16) * (26-20))
A = 130.9961832
A = 130 u ^ 2
Answer:
The area of triangle ABC is:
C.
130 u ^ 2
SEE ATTACHED IMAGE.
Using Heron's Formula we can find the area of the triangle.
A = root (s (s-a) * (s-b) * (s-c))
s = (a + b + c) / 2
Where,
s: semi-perimeter
a, b, c: sides of the triangle
Substituting values:
s = (15 + 16 + 20) / 2
s = 25.5
s = 26
The area will be:
A = root (26 * (26-15) * (26-16) * (26-20))
A = 130.9961832
A = 130 u ^ 2
Answer:
The area of triangle ABC is:
C.
130 u ^ 2
Answer:
- r = √(3V/(πh))
- r ≈ 1.78 cm
Step-by-step explanation:
Multiply by the inverse of the coefficient of r², then take the square root.
For V=50, h=15, this is ...
