1/2*bh=75
1/2*bb=75
b^2=150
b=rt(150)=12.2
not sure what you mean by link, but 12.2 is approximately the length you're seeking
The value of 7 in 7,283 is 10 times the value of 7 in 2,783
Answer:
![f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7Bx-4%7D%20%2C%20g%28x%29%3D6x%5E%7B2%7D%5Ctextrm%7B%20or%20%7Df%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D%2Cg%28x%29%3D6x%5E%7B2%7D%20-4)
Step-by-step explanation:
Given:
The function, ![H(x)=\sqrt[3]{6x^{2}-4}](https://tex.z-dn.net/?f=H%28x%29%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D)
Solution 1:
Let ![f(x)=\sqrt[3]{x}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D)
If ![f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}](https://tex.z-dn.net/?f=f%28g%28x%29%29%3DH%28x%29%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D)
, then,
![\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bg%28x%29%7D%20%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D%5C%5Cg%28x%29%3D6x%5E%7B2%7D-4)
Solution 2:
Let ![f(x)=\sqrt[3]{x-4}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7Bx-4%7D)
. Then,
![f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}](https://tex.z-dn.net/?f=f%28g%28x%29%29%3DH%28x%29%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D%5C%5C%5Csqrt%5B3%5D%7Bg%28x%29-4%7D%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D%20%5C%5Cg%28x%29-4%3D6x%5E%7B2%7D-4%5C%5Cg%28x%29%3D6x%5E%7B2%7D)
Similarly, there can be many solutions.
Answer:
$8016.00
Step-by-step explanation:
I = 6000 × 0.042 × 8 = 2016
I = $ 2,016.00
$2,016.00 + $6,000.00 = 8,016.00
(x - 2)² + (y - 6)² = 4
You can be certain about one thing just by looking at the equation (2, 6) is the center, so, obviously the circle isnt going through this point
Since the radius is 2 if we don't move from y = 6 we have points in (0, 6) and (4, 6)
So alternative b.
To be more certain just subs the point in the x and y, if its equal, it pass through
(x - 2)² + (y - 6)² = 4
To point (4, 6)
(4 - 2)² + (6 - 6)² = 4
(2²) + 0² = 4
4 = 4
Thats right
