*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
it's plural so axes*
and in pilar coordinates, the axes are circular so they never intersect but the centre point is still called origin.
To determine the maximum and minimum temperatures at which nitrogen remains a liquid, use the equation |x - (-333.32)| = 12.78, where x is equal to the maximum and minimum temperatures. Using this equation, the maximum temperature is -320.54 degrees Fahrenheit while the minimum is -346.10 degrees Fahrenheit.
So every 5 inches in height, there are 32 marbles. You want to know how many marbles there are per 18 inches in height.

To get from 5 to 18, you multiply by 3.6.
Do the same thing to the numerator; 32 × 3.6 = 115.2.
Since 0.2 marbles is no marbles, you round down.
There are 115 marbles in the box.
♡ Hope this helps! ♡
❀ 0ranges ❀
Answer:
the question is incomplete, the complete question is
"Finding Derivatives Implicity In Exercise,Find dy/dx implicity .
"
Answer : 
Step-by-step explanation:
From the expression
" y is define as an implicit function of x, hence we differentiate each term of the equation with respect to x.
we arrive at

for the expression
we differentiate using the product rule, also since y^2 is a function of y which itself is a function of x, we have
.
if we make dy/dx subject of formula we arrive at
