Answer:
- <u><em>The solution to f(x) = s(x) is x = 2012. </em></u>
Explanation:
<u>Rewrite the table and the choices for better understanding:</u>
<em>Enrollment at a Technical School </em>
Year (x) First Year f(x) Second Year s(x)
2009 785 756
2010 740 785
2011 690 710
2012 732 732
2013 781 755
Which of the following statements is true based on the data in the table?
- The solution to f(x) = s(x) is x = 2012.
- The solution to f(x) = s(x) is x = 732.
- The solution to f(x) = s(x) is x = 2011.
- The solution to f(x) = s(x) is x = 710.
<h2>Solution</h2>
The question requires to find which of the options represents the solution to f(x) = s(x).
That means that you must find the year (value of x) for which the two functions, the enrollment the first year, f(x), and the enrollment the second year s(x), are equal.
The table shows that the values of f(x) and s(x) are equal to 732 (students enrolled) in the year 2012,<em> x = 2012. </em>
Thus, the correct choice is the third one:
- The solution to f(x) = s(x) is x = 2012.
Answer:
No solutions
Step-by-step explanation:
Solve for x the system of two inequalities:

First, solve each inequality separately:
1.

2.

Now, we get

and

There are no common values for x, for which both inequalities hold, so the system of two inequalities has no solutions.
After plotting the quadrilateral in a Cartesian plane, you can see that it is not a particular quadrilateral. Hence, you need to divide it into two triangles. Let's take ABC and ADC.
The area of a triangle with vertices known is given by the matrix
M =
![\left[\begin{array}{ccc} x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20x_%7B1%7D%26y_%7B1%7D%261%5C%5Cx_%7B2%7D%26y_%7B2%7D%261%5C%5Cx_%7B3%7D%26y_%7B3%7D%261%5Cend%7Barray%7D%5Cright%5D%20)
Area = 1/2· | det(M) |
= 1/2· | x₁·y₂ - x₂·y₁ + x₂·y₃ - x₃·y₂ + x₃·y₁ - x₁·y₃ |
= 1/2· | x₁·(y₂ - y₃) + x₂·(y₃ - y₁) + x₃·(y₁ - y₂) |
Therefore, the area of ABC will be:
A(ABC) = 1/2· | (-5)·(-5 - (-6)) + (-4)·(-6 - 7) + (-1)·(7 - (-5)) |
= 1/2· | -5·(1) - 4·(-13) - 1·(12) |
= 1/2 | 35 |
= 35/2
Similarly, the area of ADC will be:
A(ABC) = 1/2· | (-5)·(5 - (-6)) + (4)·(-6 - 7) + (-1)·(7 - 5) |
= 1/2· | -5·(11) + 4·(-13) - 1·(2) |
= 1/2 | -109 |
<span> = 109/2</span>
The total area of the quadrilateral will be the sum of the areas of the two triangles:
A(ABCD) = A(ABC) + A(ADC)
= 35/2 + 109/2
= 72
399.6 square meters
7•8/2=28
28•3=84
8•6.9/2=27.6
(12•8)•3=288
84+27.6+288=399.6
Answer:
0.504m
Step-by-step explanation:
First, we would multiply 0.4cm by 126 watermelon seeds to find the length of the line in cm.
0.4*126
= 50.4cm
Therefore, the line of watermelon seeds stretches 50.4cm.
Now, because the question wants us to type the answer in meters, we must convert 50.4cm into meters. We do this by dividing 50.4 by 100 since 100cm makes up 1 meter.
50.4/100
= 0.504m
Therefore, the line of watermelon seeds stretches 0.504m.
I hope this helps!