The correct name for the line shown is the answer choice C.

It can be A because the picture is a line not a ray and it cant be B because the picture shown is not a segment

Hope this helps :)

4 0

3 years ago

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HELP ASAPPPP!!! Question in pic. Find the value of x

erik [133]

Answer:

x = 14

Step-by-step explanation:

30 = 3x - 12 (vertically opp)

30 + 12 = 3x

42 = 3x

42÷3 = x

14 = x

5 0

3 years ago

Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?

9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

$f(x)=\frac{(x-1)(x^2-4)}{x^2-a}$

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that $x=\sqrt{a}$ will annulate the denominator, i.e

$(\sqrt{a})^2-a = a-a = 0$ . But, if a = 1 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}$

so, the value $x=\sqrt{a} = \sqrt{1} = 1$ won't annulate the denominator.

Now, for a = 4 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1$

so, the value $x=\sqrt{a} = \sqrt{4} = 2$ won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0

3 years ago