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sergiy2304 [10]
3 years ago
5

The first photo in the collage is what I should make a copy of, anyone has any knowledge on how to add the numbers apart of the

spreadsheet and the letters? ​

Mathematics
2 answers:
krok68 [10]3 years ago
8 0
You can make a new table and add the amounts and copy the pre existing amounts or just add a new bar but I think it would be best if you just make it again
raketka [301]3 years ago
3 0

Answer:

uh no new copy

Step-by-steYou can make a new table to add the amounts and copy the pre-existing amounts or just add a new bar, but I guess it will be better if you just do it again.

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if 1 is added to the denominator of a fraction the fraction becomes 1/2.if 1is added to the numerator of the fraction ,the fract
saw5 [17]

Tricky but not crazy hard. You could use th x/y method where you turn it into a cross multiplication fiasco.

or you can try your guesses through what ever equals 1/2 when you add one to the denominator. the answer you are looking for is 2/3

2/3 + 1 for the numerator = 1.

2/3 + 1 for the denominator = 1/2.

6 0
3 years ago
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What is the slope of this line?
lianna [129]
The slope would be 0.
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Thang decided to borrow ​$4000 from his local bank to help pay for a car. His loan was for 3 years at a simple interest rate of
BlackZzzverrR [31]

Answer:

840

Step-by-step explanation:

I=PRT

I =4000 x 0.07 x 3

I=840

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3 years ago
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Which of the following subsets of ℝ3×3 are subspaces of ℝ3×3? A. The 3×3 matrices whose entries are all greater than or equal to
Debora [2.8K]

Answer:

A. It is NOT a subspace of R^3x3

B. It IS a subspace of R^3x3

C. It IS a subspace of R^3x3

D. It is NOT a subspace of R^3x3

E  It is NOT a subspace of R^3x3

F.  It IS a subspace of R^3x3

Step-by-step explanation:

A way to show that a set is not a subspace, it´s enough to show that some properties of the definition of a vector spaces does not apply in that set or that operations under that set are not closed (we can get out of the set with linear combinations of elements in the set).

A. For definition of subspace, we know that every element has to have an additive inverse, but in set "A" (The 3×3 matrices whose entries are all greater than or equal to 0 ) every entry is greater than or equal to zero. In this set, there´s no additive inverse with the usual sum in R^3x3.

If sufficient to prove a set is a subspace showing that zero is in the set, there are additive inverses and that operations (sum and scalar multiplication) are closed in that set.

B.  Notice that the matrix 0 is in "B" (The 3×3 matrices A such that the vector (276) is in the kernel of A), also notice if A(276)=0 then -A(276)=0 so every additive inverse (of an element in "B") belongs to "B".

Now we just have to prove that operations are closed in "B". Let X,Y matrices in set "B" and let z a scalar from the field. We are going to show that:

zX+Y ∈ B

For definition of set B:

X(276)=0 and Y(276)=0

So for zX+Y:

(zX+Y)(276)=zX(276)+Y(276)=z(0)+(0)

(zX+Y)(276)=0

So (276) is in the kernel of zX+Y, i.e (zX+Y) ∈ B.

We conclude "B" (with usual sum and scalar product of matrices) is a subspace of R^3x3

C. Notice the matrix 0 ∈ "C" (The diagonal 3×3 matrices) and there are all the additive inverse of the elements in "C". With the usual sum and scalar product, if the only zero entries are above and under the diagonal, it´ll stay like that no matter what linear combination we do because sum of matrices is entry by entry, and for every entry above or under the diagonal the sum and scalar product of two elements is going to be 0 in the same entries under and above the diagonal. "C" is a subspace

D.  In set "D" (The non-invertible 3×3 matrices) it´s necessary to show that the sum is not closed:

Consider the following matrices and their sum:

X=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right]\\ Y=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&1\end{array}\right]

X+Y=I

We showed that sum is not closed in "C", so "C" is not a subspace of R^3x3

E. The definition of a reduced row-echelon matrix requires that the first element of a row must be 1, but with sum and scalar multiplication is easy to show that these pivot could easily change its value. So the set "E" is not closed under the usual operations under R^3x3.

F. The argument is similar to part C. No matter what linear combination we do, the last row is always going to be zero (with the usual operations in R^3x3). 0 ∈ "F" (The 3×3 matrices with all zeros in the third row) and all additive inverses (for an element in "F") is in "F", we affirm that "F" is a subspace of R^3x3

5 0
3 years ago
collins middle school has 264 sixth grade students. if the sixth grade is 40% of the total school, how many students are in the
mixer [17]

Answer:

660

Step-by-step explanation:

y x .40= 264

solve for y

5 0
3 years ago
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