Answer:
Ki
Step-by-step explanation:
Answer:
-2/3
Step-by-step explanation:
Log125(1/25)=x
Raise each side to the base of 125
125^Log125(1/25)=125^x
1/25 = 125^x
Rewrite 25 as a power of 5 and 125 as a power of 5
1 / 5^2 = 5^3^x
The if power is in the denominator, we can bring it to the numerator by making it negative
5^-2 = 5^3^x
We know that a^b^c = a^(b*c)
5^-2 = 5^(3*x)
Since the bases are the same, the exponents are the same
-2 = 3x
Divide by 3
-2/3 = 3x/3
-2/3 =x
b, but I'm not sure. this could very easily be incorrect. you may need to look this up on google first.
Answer:
No, because it fails the vertical line test ⇒ B
Step-by-step explanation:
To check if the graph represents a function or not, use the vertical line test
<em>Vertical line test:</em> <em>Draw a vertical line to cuts the graph in different positions, </em>
- <em>if the line cuts the graph at just </em><em>one point in all positions</em><em>, then the graph </em><em>represents a function</em>
- <em>if the line cuts the graph at </em><em>more than one point</em><em> </em><em>in any position</em><em>, then the graph </em><em>does not represent a function </em>
In the given figure
→ Draw vertical line passes through points 2, 6, 7 to cuts the graph
∵ The vertical line at x = 2 cuts the graph at two points
∵ The vertical line at x = 6 cuts the graph at two points
∵ The vertical line at x = 7 cuts the graph at one point
→ That means the vertical line cuts the graph at more than 1 point
in some positions
∴ The graph does not represent a function because it fails the vertical
line test
Use Photomath it’s so much faster!<3
