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devlian [24]
3 years ago
5

During a winter sale,skirts are 15$ each and trousers and 20$ each. The skirts and trousers Shelly can buy for 100$

Mathematics
1 answer:
juin [17]3 years ago
8 0

Answer:

plz i need help

Step-by-step explanation:

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Brums [2.3K]

What is the problem?

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Which term best describes the graph of y = x - 1
velikii [3]
A (it’s a slight increase)
if you have a TI-84 calculator you could solve this problem in seconds.
3 0
3 years ago
A denotes an mn matrix. Determine whether the statement is true or false. Justify your answer. The row space of AT is the same a
ziro4ka [17]

Answer: true

Step-by-step explanation:

For an m*n matrix, the column space of A will be a space formed by the lineal combination of all the columns of A.

column space = a1*c1 + a2*c2 + ...

                       

where a1, a2, ... are scalars, and c1 is the vector of column 1.

Then we should write:

Column space = a1*(A₁₁, A₂₁, A₃₁, ...) + a2*(A₁₂, A₂₂, A₃₂, ...) + ...

Now, the transpose is defined as:

[At]₁₃ = A₃₁

Here i used the element with subindex 3 and 1, but is the same for every subindex.

Notice that if A is m*n, then [At] is n*m

Now, the row space of [At] will be, same as before.

Row space = b1*r1 + b2*r2 + ...

Where b1, b2, ... are scalars and the r's are the vector of each row.

                   = b1*( [At]₁₁ , [At]₁₂, [At]₁₃, ...) + b2*([At]₂₁, [At]₂₂, [At]₂₃, ...) + ...

Now we replace each term of the transpose by the associated element in the original matrix.

                   = b1*( A₁₁, A₂₁, A₃₁, ...) + b2*(A₂₁, A₂₂, ...) + ....

If we take:

b1 = a1, b2 = a2, b3 = a3, ...

We will have that the row space of [At] is the same as the column space of A.

4 0
2 years ago
The question is given in the picture.
lorasvet [3.4K]
This is a really interesting question! One thing that we can notice right off the bat is that each of the circles has the same amount of area swept out of it - namely, the amount swept out by one of the interior angles of the hexagon. Let’s call that interior angle θ. We know that the amount of area swept out in the circle is proportional to the angle swept out - mathematically

θ/360 = a/A

Where “a” is the area swept out by θ, and A is the area of the whole circle, which, given a radius of r, is πr^2. Substituting this in, we have

θ/360 = a/(πr^2)

Solving for “a”:

a = π(r^2)θ/360

So, we have the formula for the area of one of those sectors; all we need to do now is find θ and multiply our result by 6, since we have 6 circles. We can preempt this but just multiplying both sides of the formula by 6:

6a = 6π(r^2)θ/360

Which simplifies to

6a = π(r^2)θ/60

Now, how do we find θ? Let’s look first at the exterior angles of a hexagon. Imagine if you were taking a walk around a hexagon. At each corner, you turn some angle and keep walking. You make 6 turns in all, and in the end, you find yourself right back at the same place you started; you turned 360 degrees in total. On a regular hexagon, you’d turn by the same angle at each corner, which means that each of the six turns is 360/6 = 60 degrees. Since each interior and exterior angle pair up to make 180 degrees (a straight line), we can simply subtract that exterior angle from 180 to find θ, obtaining an angle of 180 - 60 = 120 degrees.

Finally, we substitute θ into our earlier formula to find that

6a = π(r^2)120/60

Or

6a = 2πr^2

So, the area of all six sectors is 2πr^2, or the area of two circles with radii r.
5 0
3 years ago
I need help with easy math please
Ivahew [28]

Answer:

1. 16

2. 12

3. 0

4. 17

5. 16

6. 4

7. 32

8. 36

9. 1/18

Step-by-step explanation:

7 0
3 years ago
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