Sorry for all the questions!!

I need to find the formula for these shaded squares

vaieri [72.5K]

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

8 0

6 months ago

Need help ASAP!!!! Been thinking for a long time.

Keith_Richards [23]

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.

6 0

2 years ago

Find the area of each figure. (Or as much as u can)

Elenna [48]

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

5 0

2 years ago

If you create a matrix, C, to show the inventory at the end of July, the value of the entry represented by C22 is ?

Inga [223]

Answer:

$C_{22} = 389$

$Max(A_{31}) =376$

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:

$C = Inventory - Sales$

i.e.

$C = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right] - \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right]$

$C = \left[\begin{array}{ccc}543-102&356-78&643-97\\364-98&476-87&419-59\\376-54&903-89&409-79\end{array}\right]$

$C = \left[\begin{array}{ccc}441&278&546\\266&389&360\\322&814&330\end{array}\right]$

Item C22 means the entry at the second row and the second column.

From the matrix

$C_{22} = 389$

Solving (b): The maximum A31 possible.

From the given data, we have:

$Inventory = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right]$

$Unit\ Sales = \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right]$

From the matrices above.

A31 means entry at the 3rd row and 1st column.

So, the possible values of A31 are:

$A_{31} = 376$

and

$A_{31} = 54$

By comparison, 376 > 54

So:

$Max(A_{31}) =376$

8 0

2 years ago

Sate the equation that relates arc length to central angle

kobusy [5.1K]

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:

5 0

2 years ago