|r+3| = (r+3) when r ≥ -3 so for r ≥ -3
r + 3 ≥ 7
r ≥ 4
|r+3| = -(r+3) when r < -3 so for r < -3
-(r + 3) ≥ 7
r + 3 ≤ -7
r ≤ -10
The answer to this question is r ≤ -10, r ≥ 4
N=12-2 is the answer of this question
Answer:
-1/8
Step-by-step explanation:
lim x approaches -6 (sqrt( 10-x) -4) / (x+6)
Rationalize
(sqrt( 10-x) -4) (sqrt( 10-x) +4)
------------------- * -------------------
(x+6) (sqrt( 10-x) +4)
We know ( a-b) (a+b) = a^2 -b^2
a= ( sqrt(10-x) b = 4
(10-x) -16
-------------------
(x+6) (sqrt( 10-x) +4)
-6-x
-------------------
(x+6) (sqrt( 10-x) +4)
Factor out -1 from the numerator
-1( x+6)
-------------------
(x+6) (sqrt( 10-x) +4)
Cancel x+6 from the numerator and denominator
-1
-------------------
(sqrt( 10-x) +4)
Now take the limit
lim x approaches -6 -1/ (sqrt( 10-x) +4)
-1/ (sqrt( 10- -6) +4)
-1/ (sqrt(16) +4)
-1 /( 4+4)
-1/8
Since a rhomubs is a 4 sided quadrilateral when parallel lines intersect within we solve for the triangles that are the result of the intersecting lines m<2=10, m<3=10
Answer: Option D.
Step-by-step explanation:
A figure has a line of symmetry if we can draw a line through the figure, in such way that the line cuts the figure in exactly two equal halves.
Then any figure that can be cutin exactly two halves, is a correct answer to this question.
Then:
For a circle, the line of symmetry is the diameter of the circle.
For the square, the line of symmetry can be obtained by cutting the square with a line that is perpendicular to one of the sides, and cuts that side exactly on the midpoint.
For an equilateral triangle, the line of symmetry is the line that cuts any base in the midpoint and also passes through the opposite vertex.
Then all the figures are correct options, then the correct option is D
