All the numbers in this range can be written as
with
and
. Construct a table like so (see attached; apparently the environment for constructing tables isn't supported on this site...)
so that each entry in the table corresponds to the sum of the tens digit (row) and the ones digit (column). Now, you want to find the numbers whose digits add to perfect squares, which occurs when the sum of the digits is either of 1, 4, 9, or 16. You'll notice that this happens along some diagonals.
For each number that occupies an entire diagonal in the table, it's easy to see that that number
shows up
times in the table, so there is one instance of 1, four of 4, and nine of 9. Meanwhile, 16 shows up only twice due to the constraints of the table.
So there are 16 instances of two digit numbers between 10 and 92 whose digits add to perfect squares.
Check the picture below, that's just an example of a parabola opening upwards.
so the cost equation C(b), which is a quadratic with a positive leading term's coefficient, has the graph of a parabola like the one in the picture, so the cost goes down and down and down, reaches the vertex or namely the minimum, and then goes back up.
bearing in mind that the quantity will be on the x-axis and the cost amount is over the y-axis, what are the coordinates of the vertex of this parabola? namely, at what cost for how many bats?
![\bf \left( -\cfrac{-7.2}{2(0.06)}~~,~~390-\cfrac{(-7.2)^2}{4(0.06)} \right)\implies (60~~,~~390-216) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\textit{number of bats}}{60}~~,~~\stackrel{\textit{total cost}}{174})~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20-%5Ccfrac%7B-7.2%7D%7B2%280.06%29%7D~~%2C~~390-%5Ccfrac%7B%28-7.2%29%5E2%7D%7B4%280.06%29%7D%20%5Cright%29%5Cimplies%20%2860~~%2C~~390-216%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%28%5Cstackrel%7B%5Ctextit%7Bnumber%20of%20bats%7D%7D%7B60%7D~~%2C~~%5Cstackrel%7B%5Ctextit%7Btotal%20cost%7D%7D%7B174%7D%29~%5Chfill)
Answer: The central angle of the arc is 162 degrees.
Step-by-step explanation: The information available are as follows;
Circumference of the circle equals 10. Length of an arc equals 9/2. The circumference of a circle is given as;
Circumference = 2Pi x r
That means 2Pi x r = 10.
Also the length of an arc along the same circle is 9/2. Length of an arc is calculated as;
Length of arc = (X/360) x 2Pi x r
Where X is the central angle of the arc
That means;
9/2 = (X/360) x 2Pi x r
We can now substitute for the known values as follows
Length of an arc = (X/360) x 2Pi x r
9/2 = (X/360) x 10
9/2 = 10X/360
By cross multiplication we now have
(9 x 360)/(2 x 10) = X
3240/20 = X
162 = X
The angle at the center of the arc is 162 degrees.
The area of a trapezoid is basically the average width times the altitude, or as a formula:
Area = h ·
b 1 + b 2
2
where
b1, b2 are the lengths of each base
h is the altitude (height)
Recall that the bases are the two parallel sides of the trapezoid. The altitude (or height) of a trapezoid is the perpendicular distance between the two bases.
In the applet above, click on "freeze dimensions". As you drag any vertex, you will see that the trapezoid redraws itself keeping the height and bases constant. Notice how the area does not change in the displayed formula. The area depends only on the height and base lengths, so as you can see, there are many trapezoids with a given set of dimensions which all have the same area.
In this We need one more side congruent because one is common i.e FG. It is the line which is Dividing it into two triangles. The property used here is SAS or Side-angle-side Property.
