write an expression for the new balance using as few terms as possible. A checking account has a balance of -$126.89. A customer
1 answer:
You might be interested in
To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]
Answer:
As x tends to +∞, P tends to 0 and as x tends to -∞, P tends to 0.
Step-by-step explanation:
Let and
, where a, b, c, d, e, f, g all are positive coefficients.
So, the function f(x) has degree two and positive leading coefficient and g(x) has degree three and negative leading coefficient.
Now, let
Therefore, as x tends to +∞, P tends to 0 and as x tends to -∞, P tends to 0. (Answer)