Answer:
add my sc = bmic ava, no spaces
or add my discord = XoXo_ava
Step-by-step explanation:
urmom gfghdfhghhgfdghjgfdsghjkhgfd fuhgjk rfujgyhbm
*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*
(21)
Area of a Regular Hexagon: square units
(22)
Similar to (21)
Area = square units
(23)
For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:
Hence, area of the hexagon will be: square units
(24)
Given is the inradius of an equilateral triangle.
Substituting the value of inradius and calculating the length of the side of the equilateral triangle:
Side = 16 units
Area of equilateral triangle = square units
Answer:
2 inches by 3 inches by 6 inches and 2 inches by 2 inches by 4 inches
Step-by-step explanation:
We can start from the given line's coefficients and translate the line from the origin to the given point.
4(x -(-2)) -(y -3) = 0
4x +8 -y +3 = 0
The equation of the desired line is ...
4x -y = -11
_____
For standard form line ax+by=c, any parallel line will have only a different value of c. For c=0, the line goes through the origin (0, 0). To make it go through point (h, k) we can write it as
a(x-h) +b(y-k) = 0
which is completely equivalent to
ax +by = ah +bk
Okay, so first, the quadrants on a coordinate plane go counter clockwise.
2 | 1
__________|__________
3 | 4
(1/2, -1.8) Would be a little to the right of the y axis, and a little below the x axis.
So, your answer would be D) Quadrant IV