12/-30=8x
cross multiply
12x=-240
x=-20
You have to searh the x values that make f(x) = 0
Factoring this equation is very difficult.
You can divide by x+3 and you will find that the equation can be factored as
(x+3)(x-4)^2
That means that x = -3 and x = 4 make f(x) = 0
Then the points are (-3,0) and (4,0).
Answer:
N = 920(1+0.03)^4t
Step-by-step explanation:
According to the given statement a car repair center services 920 cars in 2012. The number of cars serviced increases quarterly at a rate of 12% per year after 2012.
Rate is 12 % annually
rate in quarterly = 12/4= 3%
We will apply the compound interest equation:
N=P( 1+r/n)^nt
N= ending number of cars serviced.
P= the number of cars serviced in 2012,
r = interest rate
n = the number of compoundings per year
t= total number of years.
Number of compoundings for t years = n*t = 4t
Initial number of cars serviced=920
The quarterly rate of growth = n=4
r = 3%
The growth rate = 1.03
Compound period multiplied by number of years = 920(1.03)^4t
Thus N = 920(1+0.03)^4t
N = number of cars serviced after t years...
<h2>
Answer:</h2>
The ratio of the area of region R to the area of region S is:
![\dfrac{24}{25}](https://tex.z-dn.net/?f=%5Cdfrac%7B24%7D%7B25%7D)
<h2>
Step-by-step explanation:</h2>
The sides of R are in the ratio : 2:3
Let the length of R be: 2x
and the width of R be: 3x
i.e. The perimeter of R is given by:
![Perimeter\ of\ R=2(2x+3x)](https://tex.z-dn.net/?f=Perimeter%5C%20of%5C%20R%3D2%282x%2B3x%29)
( Since, the perimeter of a rectangle with length L and breadth or width B is given by:
)
Hence, we get:
![Perimeter\ of\ R=2(5x)](https://tex.z-dn.net/?f=Perimeter%5C%20of%5C%20R%3D2%285x%29)
i.e.
![Perimeter\ of\ R=10x](https://tex.z-dn.net/?f=Perimeter%5C%20of%5C%20R%3D10x)
Also, let " s " denote the side of the square region.
We know that the perimeter of a square with side " s " is given by:
![\text{Perimeter\ of\ square}=4s](https://tex.z-dn.net/?f=%5Ctext%7BPerimeter%5C%20of%5C%20square%7D%3D4s)
Now, it is given that:
The perimeters of square region S and rectangular region R are equal.
i.e.
![4s=10x\\\\i.e.\\\\s=\dfrac{10x}{4}\\\\s=\dfrac{5x}{2}](https://tex.z-dn.net/?f=4s%3D10x%5C%5C%5C%5Ci.e.%5C%5C%5C%5Cs%3D%5Cdfrac%7B10x%7D%7B4%7D%5C%5C%5C%5Cs%3D%5Cdfrac%7B5x%7D%7B2%7D)
Now, we know that the area of a square is given by:
![\text{Area\ of\ square}=s^2](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20square%7D%3Ds%5E2)
and
![\text{Area\ of\ Rectangle}=L\times B](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20Rectangle%7D%3DL%5Ctimes%20B)
Hence, we get:
![\text{Area\ of\ square}=(\dfrac{5x}{2})^2=\dfrac{25x^2}{4}](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20square%7D%3D%28%5Cdfrac%7B5x%7D%7B2%7D%29%5E2%3D%5Cdfrac%7B25x%5E2%7D%7B4%7D)
and
![\text{Area\ of\ Rectangle}=2x\times 3x](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20Rectangle%7D%3D2x%5Ctimes%203x)
i.e.
![\text{Area\ of\ Rectangle}=6x^2](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20Rectangle%7D%3D6x%5E2)
Hence,
Ratio of the area of region R to the area of region S is:
![=\dfrac{6x^2}{\dfrac{25x^2}{4}}\\\\=\dfrac{6x^2\times 4}{25x^2}\\\\=\dfrac{24}{25}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B6x%5E2%7D%7B%5Cdfrac%7B25x%5E2%7D%7B4%7D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B6x%5E2%5Ctimes%204%7D%7B25x%5E2%7D%5C%5C%5C%5C%3D%5Cdfrac%7B24%7D%7B25%7D)