Answer:
y= -50x + 400
Step-by-step explanation:
-50 miles from the place which means they are going further and further away so it is negative. And its 400 because y=mx+b the b is 400 because thats where they started. Also it said it was correct on A-p-e-x.
Answer:
Step-by-step explanation:
probability ∈[0,1]
or probability ≥0
and ≤1
so a,b,g cannot be the probability of an event.
there is an error because -9 x 1 = -9 not -1. But I would assume that you multiply -9×-1 and then -9×-2 ect.
We can use quadratic formula to determine the roots of the given quadratic equation.
The quadratic formula is:

b = coefficient of x term = 5
a = coefficient of squared term = 1
c = constant term = 7
Using the values, we get:
So, the correct answer to this question are option E and F
Answer:
s(x) is t(x) ...
- horizontally compressed by a factor of 2,
- reflected across the y-axis, and
- translated downward 5 units.
Domain and Range
- t(x) has a domain of x ≤ 0, and a range of y ≥ 0.
- s(x) has a domain of x ≥ 0, and a range of y ≥ -5.
Step-by-step explanation:
t(x) is the square root function reflected across the y-axis and compressed horizontally by a factor of 2. That is, in f(x) = √x, the x has been replaced by -2x.
s(x) has the function t(x) <em>reflected back across the y-axis</em> and compressed horizontally by another factor of 2. It is also <em>translated downward by 5 units</em>, so that its origin (vertex) is at (0, -5).
_____
The graph shows you the domain and range of s(x). The domain is all numbers to the right of x=0, including x=0. That is ...
domain: x ≥ 0
The range is all numbers -5 or above:
range: y ≥ -5
___
For t(x), the argument of the square root function must not be negative, which means the value of x cannot be positive.
domain: x ≤ 0
For non-negative values of radicand, the t(x) function will have non-negative values. So, the range is ...
range: y ≥ 0
_____
<em>Comment on solving problems like this</em>
Your graphing calculator can be your friend.