Hi there,
Your question is "twice a number (x) is fifty"
Recall how a number is represented by x.
Twice a number would be represented as 2x.
Therefore, to represent that twice a number is equal to 50, that would result in the expression: 
If you're looking to solve this equation, it's often best to think about the "reverse" of the current equation.
Notice how 2 is multiplying the x.
To isolate x, divide that side by 2.
However, recall that whatever is done to one side of an equation must also be done to the other side of the equation. Therefore, you will need to divide both sides by 2.



Equation: 
Solving for x: 25
Hope that helps!
Answer:
We conclude that option 'C' i.e. 1/3x+1=2x-4 is the correct option.
Step-by-step explanation:
From the given graph, it is clear that the two lines meet at the point (3, 2).
In other words,
Thus,
The point of intersection between two lines is:
(x, y) = (3, 2)
Here:
x = 3 is the value of the x-coordinate
y = 2 is the value of the y-coordinate
Now, checking the equation i.e. 1/3x+1=2x-4 to determine whether the equation contains the correct value of x-coordinate or not.

Subtract 1 from both sides

Simplify

subtract 2x from both sides

Simplify

Multiply both sides by 5

Simplify

Divide both sides by -5

Simplify

Therefore, the value of x = 3
Conclusion:
We conclude that option 'C' i.e. 1/3x+1=2x-4 is the correct option.
Answer:
option A and B

and

Step-by-step explanation:
we have

The formula to solve a quadratic equation of the form
is equal to

in this problem we have

so

substitute in the formula


so

and

For a function to have a derivative in a point has to be continuous at the point, that is, it has to be defined at that point, other wise the derivative would be meaningless.