A=p(1-r)^n
A=24,000×(1−0.09)^(5)
A=14,976.77
<h3>
Answer: Choice B</h3>
The parent function y = sqrt(x) has the point (0,0) on it. When we subtract 3, we get y = sqrt(x)-3 which moves the point (0,0) three units down to get to (0,-3)
So (0,-3) is one point on y = sqrt(x)-3
For simple integer factors as this one has, you want to find two values for the quadratic in the form ax^2+bx+c. Let the two values be j and k. These two values must satisfy two conditions.
jk=ac=10 and j+k=b=11, so j and k must be 1 and 10.
Now replace bx with jx and kx...
2x^2+x+10x+5 now factor 1st and 2nd pair of terms.
x(2x+1)+5(2x+1)
(x+5)(2x+1)
Answer:
Step-by-step explanation:
16 ae =x
X^1/2 ( x + 3) (x - 6)
= x^1/2 ( x^2 - 3x - 18)
= x^5/2 - 3x^3/2 - 18x^1/2
= (sqrt x)^5 - 3 (sqrt x)^3 - 18x^1/2
Option 1
