Answer (4):
a=ub/k because
u=ak/b (starting equation)
ub=ak (multiply b to both sides)
a=ub/k (divide both sides by k)
Answer (6)
x=g+c-y because
g=x-c+y (starting equation)
x+y=g+c (add c to both sides)
x=g+c-y (subtract y from both sides)
The set of side measurements that could be used to form a right triangle is 3, 4 and 5 option (D) is correct.
What is a right-angle triangle?
It is a triangle in which one of the angles is 90 degrees and the other two are sharp angles. The sides of a right-angled triangle are known as the hypotenuse, perpendicular, and base.
It is given that:
The side length of the triangle is shown in the option:
As we know,
Pythagoras' theorem is the square of the hypotenuse in a right-angled triangle is equal to the sum of the squares of the other two sides.
Using Pythagoras' theorem:
From option(D):
3, 4 and 5
(3)² + (4)² = 5²
9 + 16 = 25 (True)
Thus, the set of side measurements that could be used to form a right triangle is 3, 4 and 5 option (D) is correct.
Learn more about the right-angle triangle here:
brainly.com/question/3770177
#SPJ1
The answer is A.) 3x-7y and -7y+3x
This is due to "3x" implies the value is positive, or "+3x", and -7y remains -7y.
So, we have 3x-7y and -7y+3x.
I hope this helps!
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.
