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: B. 53
Step-by-step explanation:
We know that
<span>The formula for the area of a quadrilateral with perpendicular diagonals is
</span>A=(0.5)*(D1*D2)
A=58 in²
D1=AC=14.5 in
D2=BD=?
so
D2=2*A/D1-----------> 2*58/14.5-----------> D2=8 in
BD=D2=8 in
the answer is
BD=8 in
OK. I did it. Now let's see if I can go through it without
getting too complicated.
I think the key to the whole thing is this fact:
A radius drawn perpendicular to a chord bisects the chord.
That tells us several things:
-- OM bisects AB.
'M' is the midpoint of AB.
AM is half of AB.
-- ON bisects AC.
'N' is the midpoint of AC.
AN is half of AC.
-- Since AC is half of AB,
AN is half of AM.
a = b/2
Now look at the right triangle inside the rectangle.
'r' is the hypotenuse, so
a² + b² = r²
But a = b/2, so (b/2)² + b² = r²
(b/2)² = b²/4 b²/4 + b² = r²
Multiply each side by 4: b² + 4b² = 4r²
- - - - - - - - - - -
0 + 5b² = 4r²
Repeat the
original equation: a² + b² = r²
Subtract the last
two equations: -a² + 4b² = 3r²
Add a² to each side: 4b² = a² + 3r² . <=== ! ! !
