There are two ways you can write it:
1. 30 + 6 + (1 × 1/10) + (3 × 1/100)
2. 30 + 6 + 0.1 + 0.03
Either is correct :)
Answer:
The last listed functional expression:
Step-by-step explanation:
It is important to notice that the two linear expressions that render such graph are parallel lines (same slope), and that the one valid for the left part of the domain, crosses the y-axis at the point (0,2), that is y = 2 when x = 0. On the other hand, if you prolong the line that describes the right hand side of the domain, that line will cross the y axis at a lower position than the previous one (0,1), that is y=1 when x = 0. This info gives us what the y-intercepts of the equations should be (the constant number that adds to the term in x in the equations: in the left section of the graph, the equation should have "x+2", while for the right section of the graph, the equation should have x+1.
It is also important to understand that the "solid" dot that is located in the region where the domain changes, (x=2) belongs to the domain on the right hand side of the graph, So, we are looking for a function definition that contains 
for the function, for the domain: 
.
Such definition is the one given last (bottom right) in your answer options.
If you've started pre-calculus, then you know that the derivative of h(t)
is zero where h(t) is maximum.
The derivative is h'(t) = -32 t + 96 .
At the maximum ... h'(t) = 0
32 t = 96 sec
t = 3 sec .
___________________________________________
If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.
In that case, the question GIVES you the maximum height.
Just write it in place of h(t), then solve the quadratic equation
and find out what 't' must be at that height.
150 ft = -16 t² + 96 t + 6
Subtract 150ft from each side: -16t² + 96t - 144 = 0 .
Before you attack that, you can divide each side by -16,
making it a lot easier to handle:
t² - 6t + 9 = 0
I'm sure you can run with that equation now and solve it.
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.
(Funny how the two widely different methods lead to the same answer.)
The answer is from AL2006
Hello,
one of two<span> fixed points inside an ellipse from which the </span>sum<span> of the </span>distances<span> to any point on the ellipse is constant. </span>
Answer:19.625
Step-by-step explanation:
