Answer:
Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0. Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote. Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.
Step-by-step explanation: Correct me if i'm wrong lol.
Sí, porque un triángulo equilátero tiene 3 lados congruentes y 3 ángulos congruentes.
Answer:
Classifications would include
Rational; Fraction; Can be turned into Decimal; Positive
Those would be classifications
for the area of the smaller semicircles
area = 1/2 x 3.142 x 5 x 5
area = 39.275cm²
for the larger circle
area = 1/2 x 3.142 x 10x 10
area= 157.1cm²
but since part of it is out and is replaced again
..total area of shaded part is 157.1cm² + 39.275cm² - 39.275cm²
area =157.1cm²
1.Drawn a straight line AB=7 cm with the help of ruler.
2.With the help of compass drawn an arc from A and at the point where it cuts AB from that point made another arc drawn an arc cutting the previous arc.
3.From A drawn a straight line joining the arc and extend it to M.
4.With the help of ruler measured 5 cm and mark it as AC.
5.Joined BC and we get the required triangle.
6.From C drawn an arc and make it cut on AC and BC and from the point it cuts AC and BC drawn arc cutting each other and extend a line from point C extend a line to the point point of intersection of two arc.
7.Similarly we do for A and the point where the two line intersect denoted as O.
8.Made a perpendicular from O on AB this perpendicular will be radius and taking O as centre we draw a circle this is our incircle.
9.And AN is our locus of points equidistant from two lines AB and AC.
We need to construct a circle inscribed in triangle that is incircle it can be done by making angle bisector of two sides the point where it intersect will be incentre. The centre of required circle.
The angle bisector is the locus where points are equidistant from two sides.