Solve the following equation. Then place the correct number in the box provided.
2 answers:
You might be interested in
Step 1
<u>Find the equation in function notation</u>
Let
h-------> the number of hours
y-------> the function for the walker fee in dollars
we know that
the hourly rate is
in this linear equation
the independent variable is the variable h
the dependent variable is the variable y
<u>Convert to function notation</u>
Let
Step 2
Find how much does the dog walker charge for a 45 minute walk
Convert the time in hours
substitute in the equation
For
therefore
<u>the answer is</u>
the dog walker charge for a 45 minute walk
<u>Statements</u>
Step 3
<u>The dependent variable is the number of hours true or false</u>
The statement is false,
because the independent variable is the number of hours and the dependent variable is the walker fee in dollars
Step 4
<u>The function for the walker fee is f(h)= 30h+6 true or false</u>
The statement is True --------> see the Step 1
Step 5
<u>The dog walker charges $22.5 for a 45 min walk true or false</u>
The statement is False
Because the dog walker charges for a 45 minute walk
Answer:
a) No. t < 0 is not part of the useful domain of the function
b) 2.0 seconds
Step-by-step explanation:
a) A graph of the function is shown below. It shows t-intercepts at t=-0.25 and t=2.0. We presume that t is measured forward from some event such as the ball being thrown or hit. The model's predicted ball location has no meaning prior to that event, when values of t are negative.
__
b) It is convenient to use a graphing calculator to find the t-intercepts. Or, the equation can be solved for h=0 any of several ways algebraically. One is by factoring.
h = 0 = -16t² +28t +8 . . . . . . . . . . . . the ball hits the ground when h = 0
0 = -4(4t² -7t -2) = -4(4t +1)(t -2)
This has t-intercepts where the factors are zero, at t=-1/4 and t=2.
The ball will hit the ground after 2 seconds.
Answer:
1. A
2. D
3. D
Step-by-step explanation:
The standard form of a parabola is
..... (1)
Where, (h,k) is vertex, (h,k+p) is focus and y=k-p is directrix.
1. The directrix of a parabola is y=−8 . The focus of the parabola is (−2,−6) .
...(a)
.... (b)
On solving (a) and (b), we get k=-7 and p=1.
Put h=-2, k=-7 and p=1 in equation (1).
Therefore option A is correct.
2 The directrix of a parabola is the line y=5 . The focus of the parabola is (2,1) .
...(c)
.... (d)
On solving (c) and (d), we get k=3 and p=-2.
Put h=2, k=3 and p=-2 in equation (1).
Therefore option D is correct.
3. The focus of a parabola is (0,−2) . The directrix of the parabola is the line y=−3 .
...(e)
.... (f)
On solving (e) and (f), we get k=-2.5 and p=0.5.
Put h=0, k=-2.5 and p=0.5 in equation (1).
Therefore option D is correct.