Answer:
Step-by-step explanation:
Let x represent the number of years it will take the two colleges to have the same enrollment.
In 2000, there were 12900 students at college A, with a projected enrollment increase of 900 students per year. This means that the expected number of students at college A in x years time is
12900 + 900x
In the same year, there were 25,000 students at college B, with a projected enrollment decline of 700 students per year. This means that the expected number of students at college B in x years time is
25000 - 700x
For both colleges to have the same enrollment,
12900 + 900x = 25000 - 700x
900x + 700x = 25000 - 12900
1600x = 12100
x = 12100/1600
x = 7.56
Approximately 8 years
The year would be 2000 + 8 = 2008
Answer:
The simplified form of the expression is ![\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
Step-by-step explanation:
Given : Expression ![7\sqrt[3]{2x}-3\sqrt[3]{16x}-3\sqrt[3]{8x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-3%5Csqrt%5B3%5D%7B16x%7D-3%5Csqrt%5B3%5D%7B8x%7D)
To Simplified : The expression
Solution :
Step 1 - Write the expression
![7\sqrt[3]{2x}-3\sqrt[3]{16x}-3\sqrt[3]{8x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-3%5Csqrt%5B3%5D%7B16x%7D-3%5Csqrt%5B3%5D%7B8x%7D)
Step 2- Simplify the roots and re-write as
and 
![7\sqrt[3]{2x}-3\times2\sqrt[3]{2x}-3\times2\sqrt[3]{x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-3%5Ctimes2%5Csqrt%5B3%5D%7B2x%7D-3%5Ctimes2%5Csqrt%5B3%5D%7Bx%7D)
Step 3- Solve the multiplication
![7\sqrt[3]{2x}-6\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
Step 4- Taking
common from first two terms
![\sqrt[3]{2x}(7-6)-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D%287-6%29-6%5Csqrt%5B3%5D%7Bx%7D)
![\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
Therefore, The simplified form of the expression is ![\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
I'll denote the identity function by
. Then for any functions
with inverse
,

One important fact is that composition is associative, meaning for functions
, we have

So given

we can compose the functions on either side with
:

then apply the rules listed above:

Answer: m=1/30n+29/15
Step-by-step explanation:
Answer:
Hence when the radius is halved the area is divided by 4
2.5 inches^2
Step-by-step explanation:
Given data
Area= 10inches^2
We know that the expression for the area of a circle is given as
Area= πr^2
10= 3.142*r^2
10/3.142= r^2
r^2= 3.18
Square both sides
r= √3.18
r= 1.78 inches
Now let us half the radius and find the area of the new circle
r/2= 1.78/2
r= 0.89
Area of the new circle is
Area= 3.142*0.89^2
Area= 3.142*0.7921
Area= 2.5 inches^2