Answer:
The solution to f(x) = t(x) is x = 2010
Option 3 is true.
Step-by-step explanation:
The first-year , second-year , and third-year enrollment values for a technical school are shown in the table below.
Year (x) First Year f(x) Second Year s(x) Third Year t(x)
2009 785 756 756
2010 740 785 740
2011 690 710 781
2012 732 732 710
2013 781 755 800
Now we will check each option.
Option 1: The solution to f(x) = s(x) is x = 2,009
In year 2009, f(x)=s(x)
But 785≠756
Thus, False
Option 2: The solution to f(x) = s(x) is x = 785
x represents year, but 785 it no year
Thus, False
Option 3: The solution to f(x) = t(x) is x = 2010
In year 2010, f(x)=t(x)=740
But 740=740
Thus, True
Option 4: The solution to f(x) = t(x) is x =740
x represents year, but 740 it no year
Thus, False
Answer:
![-7ab\sqrt[3]{3ab^2}](https://tex.z-dn.net/?f=-7ab%5Csqrt%5B3%5D%7B3ab%5E2%7D)
Step-by-step explanation:
Remove perfect cubes from under the radical and combine like terms.
![2ab\sqrt[3]{192ab^2}-5\sqrt[3]{81a^4b^5}=2ab\sqrt[3]{4^3\cdot 3ab^2}-5\sqrt[3]{(3ab)^3\cdot 3ab^2}\\\\=(8ab -15ab)\sqrt[3]{3ab^2}=\boxed{-7ab\sqrt[3]{3ab^2} }](https://tex.z-dn.net/?f=2ab%5Csqrt%5B3%5D%7B192ab%5E2%7D-5%5Csqrt%5B3%5D%7B81a%5E4b%5E5%7D%3D2ab%5Csqrt%5B3%5D%7B4%5E3%5Ccdot%203ab%5E2%7D-5%5Csqrt%5B3%5D%7B%283ab%29%5E3%5Ccdot%203ab%5E2%7D%5C%5C%5C%5C%3D%288ab%20-15ab%29%5Csqrt%5B3%5D%7B3ab%5E2%7D%3D%5Cboxed%7B-7ab%5Csqrt%5B3%5D%7B3ab%5E2%7D%20%7D)
2(5x + 2)^2 = 48
divide both sides by 2
(5x + 2)^2 = 24
square root both sides to clear the power of two.
5x + 2 = + - sqrt24
subtract 2 from both sides
5x = -2 + - sqrt 24
divide both sides by 5
x = (-2 +-sqrt24)/5 or decimal answers .58 and -1.38
Start with $30.25.
Subtract $14.00 to isolate the amount of money spent on admission.
30.25-14.00= 16.25. As a group $16.25 was spent on admission.
Admission is 3.25 per person. Divide 16.25 by 3.25.
16.25/3.25=5.
5 friends went ice skating.
