$1620
calculate simple interest ( I) using I = PRT
where P is the amount borrowed, R is rate of interest and T is time in years
note that R = 6% =
= 0.06
I = $9000 × 0.06 × 3 = $1620
the answer is D because the other points dont have a line connecting them so you shoulduse the points that are exterior of square ctbe
Answer:
The Quadratic Polynomial is
2 x² +x -4=0
Using the Determinant method to find the roots of this equation

For, the Quadratic equation , ax²+ b x+c=0
(b) x²+x=0
x × (x+1)=0
x=0 ∧ x+1=0
x=0 ∧ x= -1
You can look the problem in other way
the two Quadratic polynomials are
2 x²+x-4=0, ∧ x²+x=0
x²= -x
So, 2 x²+x-4=0,
→ -2 x+x-4=0
→ -x -4=0
→x= -4
∨
x² +x² +x-4=0
x²+0-4=0→→x²+x=0
→x²=4
x=√4
x=2 ∧ x=-2
As, you will put these values into the equation, you will find that these values does not satisfy both the equations.
So, there is no solution.
You can solve these two equation graphically also.
The formula for the area of a hexagon is
![A=\frac{3\sqrt[]{3}}{2}s^2](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B3%5Csqrt%5B%5D%7B3%7D%7D%7B2%7Ds%5E2)
where 's' is the length of one side of the regular hexagon.
The side of our regular hexagon is 2 feet, therefore, its area is
![\begin{gathered} A=\frac{3\sqrt[]{3}}{2}\cdot(2)^2=6\sqrt[]{3} \\ 6\sqrt[]{3}=10.3923048454\ldots\approx10 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D%5Cfrac%7B3%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%5Ccdot%282%29%5E2%3D6%5Csqrt%5B%5D%7B3%7D%20%5C%5C%206%5Csqrt%5B%5D%7B3%7D%3D10.3923048454%5Cldots%5Capprox10%20%5Cend%7Bgathered%7D)
The exact area of the hexagon is 6√3 ft², which is approximately 10 ft².
<h3><u>The variable x is equal to (p + 11)/5.</u></h3>
P = 5x - 11
<em><u>Add 11 to both sides.</u></em>
p + 11 = 5x
<em><u>Divide both sides by 5.</u></em>
(p + 11)/5 = x