I think it's another trick question and the answer is 2000.
Answer:
The answer is 40°
Step-by-step explanation:
Solution
Now,
The Measure of major arc FD = 280°
Thus,
One complete angle measure 360°
Then
m∠FED = 360 - 180
m∠FED = 80°
Thus,
FE = DE ( the same circle Radius)
∠EFD = ∠EDF ( opposite angles to equal sides are equal)
Now
Applying angle sum property of a triangle in ΔEFD
∠EFD + ∠EDF + ∠FED = 180°
∠EFD + ∠EFD + 80 = 180
2∠EFD = 100
∠EFD = 50°
Hence, GF is tangent to the circle and the tangent always make right angles with the radius of the circle.
∠EFG = 90°
∠GFD + ∠EFD = 90°
∠GFD + 50 = 90
∠GFD = 40°
Therefore, The measure of the angle GFD is 40°
Answer:
![f(x)=4\sqrt[3]{16}^{2x}](https://tex.z-dn.net/?f=f%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D)
Step-by-step explanation:
We believe you're wanting to find a function with an equivalent base of ...
![4\sqrt[3]{4}\approx 6.3496](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7B4%7D%5Capprox%206.3496)
The functions you're looking at seem to be ...
![f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x](https://tex.z-dn.net/?f=f%28x%29%3D2%5Csqrt%5B3%5D%7B16%7D%5Ex%5Capprox%202%5Ccdot2.5198%5Ex%5C%5C%5C%5Cf%28x%29%3D2%5Csqrt%5B3%5D%7B64%7D%5Ex%3D2%5Ccdot%204%5Ex%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D%5Capprox%204%5Ccdot%206.3496%5Ex%5C%20%5Cleftarrow%5Ctext%7B%20this%20one%7D%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B64%7D%5E%7B2x%7D%3D4%5Ccdot%2016%5Ex)
The third choice seems to be the one you're looking for.
Answer:
no
Step-by-step explanation:
