The answer to your Q is 4
Answer:
The distance of the point from the origin = 9.29 units.
Step-by-step explanation:
Given point:
(7,-6)
The angle lies such that the terminal side of the angle contains the given point.
To draw the angle and find the distance from the origin to the given point.
Solution:
The terminal side of the angle is where the angle ends with the initial side being the positive side of the x-axis.
So, we can plot the point (7,-6) by moving 7 units on the x-axis horizontally and -6 units on the y-axis vertically.
We can find the distance of the point from the origin by find the hypotenuse of the triangle formed.
Applying Pythagorean theorem.



Taking square root both sides :


Thus, the distance of the point from the origin = 9.29 units.
The figure is shown below.
Answer:
is outside the circle of radius of
centered at
.
Step-by-step explanation:
Let
and
denote the center and the radius of this circle, respectively. Let
be a point in the plane.
Let
denote the Euclidean distance between point
and point
.
In other words, if
is at
while
is at
, then
.
Point
would be inside this circle if
. (In other words, the distance between
and the center of this circle is smaller than the radius of this circle.)
Point
would be on this circle if
. (In other words, the distance between
and the center of this circle is exactly equal to the radius of this circle.)
Point
would be outside this circle if
. (In other words, the distance between
and the center of this circle exceeds the radius of this circle.)
Calculate the actual distance between
and
:
.
On the other hand, notice that the radius of this circle,
, is smaller than
. Therefore, point
would be outside this circle.
No regrouping: 4.34 + 4.35
Regrouping (carrying): 4.93 + 3.76
Answer:
Step-by-step explanation:
I'm assuming you meant to type in
because you can only have removable discontinuities where there is a rational (fraction) function. Begin by factoring both the numerator and denominator to
and cancelling out like terms would have us eliminating the (x + 3). That is where there is a removable discontinuity. It leaves a hole. The other discontinuity, (x + 1) doesn't cancel out so it is a non-removable discontuinity, which is a vertical asymptote.
The removable discontinuity is at -3. There is no y value at x = -3 (remember there's only a hole here), because -3 causes the denominator to go to 0 and we all know that having a 0 in the denominator of a fraction is a big no-no!!!