For the given sequence we have the formula:
Sₙ = 1 + (n - 1)*2
The 50th square will have 99 shaded squares.
<h3 /><h3>
How many shaded squares are on the n-th square?</h3>
Here we have a sequence:
The first square has 1 shaded squares.
the second square has 3 shaded squares.
The third square has 5 shaded squares.
And so on.
Already you can see a pattern here, each next step we add 2 shaded squares, then we can write the formula:
Sₙ = 1 + (n - 1)*2
Where S is the number of shaded squares and n is the number of the figure.
Then the 50th square will have:
S₅₀ = 1+ (50 - 1)*2 1 + 49*2 = 99
Learn more about sequences:
brainly.com/question/6561461
#SPJ1
Answer:
81%
Step-by-step explanation:
First, let's set up 3 equations:
w*0.6 = x
x*1.5=y
y*0.9=z
Now, we can substitute x in the second equation for w*0.6:
w*0.6*1.5 = y
Lastly, we can substitute y in the last equation for w*0.6*1.5:
w*0.6*1.5*0.9=z
w*0.81=z
Therefore, z is 81 percent of w.
9514 1404 393
Answer:
7. 91 m²
8. 56 in²
9. 105 in²
10. 135 m²
Step-by-step explanation:
The formula for the area of a triangle is ...
A = 1/2bh
where b is the base length, and h is the height perpendicular to the base.
The formula for the area of a trapezoid is ...
A= 1/2(b1 +b2)h
where b1 and b2 are the lengths of the parallel bases, and h is the perpendicular distance between them. Note that this formula is virtually identical to the triangle formula, with the triangle 'b' being replaced by (b1+b2).
__
Fill in the numbers and do the arithmetic. The result is shown in the attachment. (We have used the triangle formula for all, with b1+b2 being used for 'b' to find the area of the trapezoids.)
7. A = 1/2(14 m)(13 m) = 91 m²
8. 56 in²
9. A = 1/2(14 +16 in)(7 in) = 105 in²
10. 135 m²
Answer:


Step-by-step explanation:
Given
See attachment for complete question
Solving (a): The entry C22
First, matrix C represents the inventory at the end of July.
The entry of C is calculated as:

i.e.
![C = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right] - \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right]](https://tex.z-dn.net/?f=C%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D543%26356%26643%5C%5C364%26476%26419%5C%5C376%26903%26409%5Cend%7Barray%7D%5Cright%5D%20%20-%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2678%2697%5C%5C98%2687%2659%5C%5C54%2689%2679%5Cend%7Barray%7D%5Cright%5D)
![C = \left[\begin{array}{ccc}543-102&356-78&643-97\\364-98&476-87&419-59\\376-54&903-89&409-79\end{array}\right]](https://tex.z-dn.net/?f=C%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D543-102%26356-78%26643-97%5C%5C364-98%26476-87%26419-59%5C%5C376-54%26903-89%26409-79%5Cend%7Barray%7D%5Cright%5D)
![C = \left[\begin{array}{ccc}441&278&546\\266&389&360\\322&814&330\end{array}\right]](https://tex.z-dn.net/?f=C%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D441%26278%26546%5C%5C266%26389%26360%5C%5C322%26814%26330%5Cend%7Barray%7D%5Cright%5D)
Item C22 means the entry at the second row and the second column.
From the matrix

Solving (b): The maximum A31 possible.
From the given data, we have:
![Inventory = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right]](https://tex.z-dn.net/?f=Inventory%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D543%26356%26643%5C%5C364%26476%26419%5C%5C376%26903%26409%5Cend%7Barray%7D%5Cright%5D)
![Unit\ Sales = \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right]](https://tex.z-dn.net/?f=Unit%5C%20Sales%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2678%2697%5C%5C98%2687%2659%5C%5C54%2689%2679%5Cend%7Barray%7D%5Cright%5D)
From the matrices above.
A31 means entry at the 3rd row and 1st column.
So, the possible values of A31 are:

and

By comparison, 376 > 54
So:
S =rO
The length of an arc is related to the radius and the central angle in the formula. s = rθ. where s is the arc length, r is the radius and θ is the angle in radians. Remember that to convert from radians to degrees, 2π radians = 360 degrees, so if φ = the angle in degrees, then the equation becomes this: