Answer:
28
Length=2(x-1)
Width=5
Area=length*width = (2(x-1))(5) = (2x-2)(5) = 10x-10
29
His reasoning is illogical because whether or not an expression has a term that is being subtracted isn't relevant; technically there are an infinite amount of ways to represent a value. Plus you can just compute the expressions and see that they're equal:
6x-2x+4 = 4x+4
4(x-1) = 4x+4
30
The two expressions are equal because when you compute the expression 4(n+3)-(3+n) , you get 3n+9:
4(n+3)-(3+n) = 4n+12-3-n = 3n+9
31
The two expressions are equal because when you compute the expression 2(2n-1), you get 4n-2.
2(2n-1) = (2)(2n)+(2)(-1)=4n-2
32
5(g+14)=(5)(g)+(5)(14)=5g+70
The expressions aren't equal as 5(g+14) equates to 5g+70 and 5g+70≠5g+14.
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.
X=43/12 If you need work shown please tell me in the comments. Also please mark me as brainliest.
F’(x)= -akxsinkx
F’’(x)= -a(kx)^2coskx
I’m really sorry if the answer turned out wrong but I tried my best!
