Answer:
The correct options are 1, 2 and 4.
Step-by-step explanation:
The given triangle is an equilateral triangle. All the sides of an equilateral triangle are same. All interior angles are equal with measure 60 degree.
The distance from the center of an equilateral triangle to the midpoint of a side is known as apothem. In the given figure letter a represents the apothem.
Point D is the midpoint of BC,

Use Pythagorean theorem in triangle COD. So, option 1 is correct.





The distance can not be negative, so the length of the apothem is approximately 2.5 cm. Option 4 is correct.
The line OC bisects the angle C.



Therefore option 2 is correct.
The perimeter of an equality triangle is

Option 3 is incorrect.

Option 5 is incorrect.
Therefore options 1, 2 and 4 are correct.
2 because in the equation of a circle it is x^2 + y^2 = r^2, r=radius so you would take the square root of 4, leaving the radius of the circle to be 2.
Answer:
Step-by-step explanation:
I need this to ekk
See photo for drawing.
Words:
Suppose a square is 100%,
50%: the square is divided into two equal parts, and one of them is shaded, therefore it is 50%.
25%: the square is divided into 4 equal parts, and one of them is shaded, therefore it is 25%.
150%: there are two squares, each divided into 2 equal parts, and 3 of them is shadowed, therefore it is 150%.
Substitute
, so that
. Then the ODE is equivalent to

which is separable as

Split the left side into partial fractions,

so that integrating both sides is trivial and we get








Given the initial condition
, we find

so that the ODE has the particular solution,
