, for a dog weighing x pounds.
f(x)=⎧⎩⎨⎪⎪2.5x if 0≤x≤203.5x if 20 <x≤50 5x if x>50
What would be the last point graphed on the first piece of the function, f(x)=2.5x
?
(___,___)
11 answers
The first piece of the piecewise function is \( f(x) = 2.5x \) and is defined for the range \( 0 \leq x \leq 20 \).
To determine the last point graphed for this piece, we need to evaluate the function at the upper limit of its domain, which is \( x = 20 \):
\[ f(20) = 2.5 \times 20 = 50 \]
Thus, the last point graphed on the first piece of the function would be:
\[ (20, 50) \]
So, the answer is \((20, 50)\).
To find the value of the car after three years using the depreciation model \( f(x) = 32,000(0.92)^x \), we need to calculate \( f(3) \).
Substituting \( x = 3 \):
\[ f(3) = 32,000(0.92)^3 \]
First, calculate \( (0.92)^3 \):
\[ (0.92)^3 = 0.92 \times 0.92 \times 0.92 \approx 0.778688 \]
Now, calculate \( f(3) \):
\[ f(3) = 32,000 \times 0.778688 \approx 24,964.00 \]
Thus, the value of the car after three years is approximately $24,964.00.
So the answer is:
$(24,964.00)$.
To determine the domain, we need to analyze the situation specified in the problem.
In this case, Marilee can build a rectangular enclosure using the fencing materials, with the perimeter given by \( P = 2l + 2w \), where \( l \) is the length and \( w \) is the width. Given that she has 40 yards of fencing, we can write the equation for the perimeter as:
\[ 2l + 2w = 40 \implies l + w = 20 \implies l = 20 - w \]
The area \( A \) of the rectangle is given by:
\[ A = l \cdot w = (20 - w) \cdot w = 20w - w^2 \]
This simplifies to the quadratic equation provided in the problem:
\[ A = -w^2 + 20w \]
To find the domain of this function, we need to consider the physical constraints on the width \( w \):
Combining these two constraints, the domain for \( w \) is:
\[ 0 \leq w \leq 20 \]
Now, you are asking about the maximum range when the maximum area is achieved. The maximum area occurs at the vertex of the quadratic equation. The \( w \)-coordinate of the vertex can be found using the formula \( w = -\frac{b}{2a} \), where \( A = aw^2 + bw + c \).
In our case, \( a = -1 \) and \( b = 20 \):
\[ w = -\frac{20}{2 \cdot -1} = 10 \]
At \( w = 10 \), we can find the corresponding area:
\[ A(10) = -10^2 + 20 \cdot 10 = -100 + 200 = 100 \text{ square yards} \]
The maximum area (range) of the garden is 100 square yards when the width is 10 yards.
Therefore, the domain we're discussing refers specifically to the width, which has a maximum of 20 yards, but the answer to your question about the maximum range achieved is correlated to:
10 yards (the width at which the maximum area is achieved).
To determine the correct pairs of \( (t, h) \) using the function \( h = -6t^2 + 33t + 6 \) for \( t = 0, 1, 2, 3, 4 \), we need to evaluate the function at each specific \( t \).
For \( t = 0 \): \[ h(0) = -6(0)^2 + 33(0) + 6 = 6 \]
For \( t = 1 \): \[ h(1) = -6(1)^2 + 33(1) + 6 = -6 + 33 + 6 = 33 \]
For \( t = 2 \): \[ h(2) = -6(2)^2 + 33(2) + 6 = -24 + 66 + 6 = 48 \]
For \( t = 3 \): \[ h(3) = -6(3)^2 + 33(3) + 6 = -54 + 99 + 6 = 51 \]
For \( t = 4 \): \[ h(4) = -6(4)^2 + 33(4) + 6 = -96 + 132 + 6 = 42 \]
Now we can summarize our results:
Thus, the correct corresponding pairs of \( (t, h) \) values are:
\[ (0,6), (1,33), (2,48), (3,51), \text{ and } (4,42) \]
The correct response from the options provided is:
(0,6), (1,33), (2,48), (3,51), and (4,42).
To analyze the meaning of the x-intercepts of the function \( A = -w^2 + 60w \), we first need to find the x-intercepts. The x-intercepts occur when the area \( A \) is equal to zero:
\[ 0 = -w^2 + 60w \]
Factoring this equation gives:
\[ 0 = w(-w + 60) \]
Setting each factor to zero, we find the x-intercepts:
The x-intercepts at \( w = 0 \) and \( w = 60 \) indicate the valid widths of the dog run:
The area function \( A = -w^2 + 60w \) is a downward-opening parabola with its maximum point occurring at \( w = 30 \) (the vertex). The area is positive between the intercepts (when \( 0 < w < 60 \)).
Based on this analysis, the correct interpretation of the x-intercepts is:
The dog run must have a width between 0 and 60 feet.
So the response is:
The dog run must have a width between 0 and 60 feet.
Situation: A local charity organizes a fundraising event where they sell tickets at a price of $5 each, and for every additional ticket sold (up to 7 tickets), the total amount raised increases linearly, with the goal of reaching $40 in donations.
Explanation: The graph models the total amount of money raised (in dollars) as a function of the number of tickets sold, showing how the total donation increases with each ticket sold.
