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BigorU [14]
2 years ago
15

Leon throws a ball off a cliff. The graph represents a relation that models the ball’s height, h, in feet over time, t, in secon

ds.
What is the domain?

{t| 0 ≤ t ≤ 3}
{t| 0 ≤ t ≤ 48}
{t| 0 ≤ t ≤ 64}
{t| all real numbers}

Mathematics
2 answers:
erik [133]2 years ago
8 0

Answer: The answer is A.


Step-by-step explanation:


otez555 [7]2 years ago
4 0

Answer:

The correct option is 1.

Step-by-step explanation:

The given graph represents a relation that models the ball’s height, h, in feet over time, t, in seconds.

The value of height and time can not be negative, there the values of h and t must be greater than 0.

The set of all possible inputs is called domain. In the graph x axis represent the domain and y-axis represent the range.

The graph is defined from t=0 to t=3.

Domain=\{t|0\leq t\leq3\}

Therefore option 1 is correct.

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Answer:

14inches farenheight

Step-by-step explanation:

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2 years ago
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A tank contains 8000 L of pure water. Brine that contains 35 g of salt per liter of water is pumped into the tank at a rate of 2
OleMash [197]

Answer:

The concentration of salt in the tank approaches 35 \mathrm{g} / \mathrm{L},

Step-by-step explanation:

Data provide in the question:

Water contained in the tank = 8000 L

Salt per litre contained in Brine = 35 g/L

Rate of pumping water into the tank = 25 L/min

Concentration of salt \lim _{t \rightarrow \infty} C(t)=\lim _{t \rightarrow \infty} \frac{35 t}{320+t}

Now,

Dividing both numerator and denominator by t, we have

\lim _{t \rightarrow \infty} \frac{\frac{1}{t} 35 t}{\frac{1}{t}(320+t)}=\lim _{t \rightarrow \infty} \frac{35}{\frac{320}{t}+1}=35

Here,

The concentration of salt in the tank approaches 35 \mathrm{g} / \mathrm{L},

3 0
2 years ago
Craig ran the first part of a race with an average speed of 8 miles per hour and biked the second part of a race with an average
lutik1710 [3]
Let the distance of the first part of the race be x, and that of the second part, 15 - x, then
x/8 + (15 - x)/20 = 1.125
5x + 2(15 - x) = 40 x 1.125
5x + 30 - 2x = 45
3x = 45 - 30 = 15
x = 15/3 = 5

Therefore, the distance of the first part of the race is 5 miles and the time is 5/8 = 0.625 hours or 37.5 minutes
The distance of the second part of the race is 15 - 5 = 10 miles and the time is 1.125 - 0.625 = 0.5 hours or 30 minutes.
4 0
2 years ago
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The height of a baseball thrown from the catcher to first base is modeled by the function h(t)=-0.08t^2+0.72t+6, where h is the
TiliK225 [7]
The answer would be c
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2 years ago
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Quadrilateral A’B’C’D’ is a dilation of quadrilateral ABCD about point P. Quadrilateral ABCD is shown. Side AB is labeled 5. Sid
Stolb23 [73]

Answer:  The correct option is (A) reduction.

Step-by-step explanation:  Given that the quadrilateral A'B'C'D' is a dilation of the quadrilateral ABCD.

As shown in the given figure, the lengths of the sides of quadrilateral ABCD are as follows:

AB = 5 units, BC = 4 units, CD = 10 units and DA = 6 units.

And, the lengths of the sides of quadrilateral A'B'C'D' are as follows:

A'B'=1\dfrac{1}{4}=\dfrac{5}{4}~\textup{units},~~B'C'=1~\textup{units},~~C'D'=2\dfrac{1}{2}=\dfrac{5}{2}~\textup{units},\\\\D'A'=1\dfrac{1}{2}=\dfrac{3}{2}~\textup{units}.

We know that the dilation will be an enlargement if the scale factor is greater than 1 and it will be a reduction if the scale factor is less than 1.

Now, the scale factor is given by

S=\dfrac{\textup{length of a side of the dilated figure}}{\textup{length of the corresponding side of the original figure}}\\\\\\\Rightarrow S=\dfrac{A'B'}{AB}=\dfrac{\frac{5}{4}}{5}=\dfrac{5}{4\times5}=\dfrac{1}{4}

Since the scale factor is less than 1, so the dilation will be a reduction.

5 0
2 years ago
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