170*3=510
510 calories in the whole bar.
Hope this helps!
A lever in like a pully becasue it can pull just as so. Hope that helped. It is what my teacher taught me.
A function f(x) has solutions if we can find a value to plug in that leads to 0. In other words, there are solutions to f(x) = 0. Another term for "solution" is "root" or "x intercept"
An exponential function may cross the x axis at one point only. Though there are plenty of cases when there are no solutions at all. For instance, in the case of f(x) = (2^x) + 10
The right hand side will never be equal to zero no matter what you plug in for x. The graph will never cross the axis.
To answer your question, yes it is possible to have an exponential equation to have no solutions.
Answer:
1. 
2. 
3. 
Step-by-step explanation:
Given information:


(1)
We need to find the value of P(s₁|I).





Therefore the value of P(s₁|I) is
.
(2)
We need to find the value of P(s₂|I).





Therefore the value of P(s₂|I) is
.
(3)
We need to find the value of P(s₃|I).





Therefore the value of P(s₃|I) is
.
Answer:
I am pretty sure we are all tired but here is your answer 195
Step-by-step explanation: