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sweet-ann [11.9K]
3 years ago
15

Is 70% of 80 the same as 80% of 70

Mathematics
2 answers:
KiRa [710]3 years ago
6 0
Yes because 70*.8=56 and 80*.7=56

White raven [17]3 years ago
4 0

No, because 80 and 70 are different numbers so there is going to be a different answer.
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Subtract the polynomials.
Mama L [17]

Answer:

B

Step-by-step explanation:

Given

(4x² - 3x - 4) - (3x² + 4x - 8) ← distribute by - 1

= 4x² - 3x - 4 - 3x² - 4x + 8 ← collect like terms

= x² - 7x + 4 → B

8 0
3 years ago
Read 2 more answers
Consider the following sets of matrices: M2(R) is the set of all 2 x 2 real matrices; GL2(R) is the subset of M2(R) with non-zer
defon

Answer:

Step-by-step explanation:

REcall the following definition of induced operation.

Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.

So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.

For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).

Case SL2(R):

Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So

\text{det(AB)} = \text{det}(A)\text{det}(B)\neq 0.

So AB is also in SL2(R).

Case GL2(R):

Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So

\text{det(AB)} = \text{det}(A)\text{det}(B)=1\cdot 1 = 1.

So AB is also in GL2(R).

With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).

7 0
3 years ago
A test requires that you answer first Part A and then either Part B or Part C. Part A consists of 4 true false questions, Part B
erma4kov [3.2K]

Answer:  374416

Step-by-step explanation:

Given : A test requires that you answer first Part A and then either Part B or Part C.

Part A consists of 4 true false questions, Part B consists of 6 multiple-choice questions with one correct answer out of five, and Part C consists of 5 multiple-choice questions with one correct answer out of six.

i.e. 2 ways to answer each question in Part A.

For 4 questions, Number of ways to answer Part A = 2^4

5 ways to answer each question in Part B.

For 6 questions, Number of ways to answer Part B = 5^6

6 ways to answer each question in Part C.

For 5 questions, Number of ways to answer Part C = 6^5

Now, the number of ways to  completed answer sheets are possible :_

2^4\times5^6+2^4\times6^5\\\\=2^4(5^6+6^5)\\\\=16(15625+7776)\\\\=16(23401)=374416

Hence, the number of ways to  completed answer sheets are possible = 374416

8 0
3 years ago
Ayayayayauauauaua help
Cerrena [4.2K]

Answer:

I cant do them all for you, but essentially every equation there is an a, plug in 10 for a. Every equation  with b, plug in 9 and every equation with c plug in 4. Then Solve/simplify

Step-by-step explanation:

5 0
3 years ago
Help with number 7!! You don’t have to do it but how do you do the problem??
musickatia [10]
80 * 5/8 will give you how many are thoroughbreds. 

80 * 5/8 = 50

Now subtract this from all of the horses. 

80 - 50 = 30. 

There are 30 quarter horses. 

OR If 5/8 of the horses are thoroughbreds, then 3/8 are quarter horses.

80 * 3/8 = 30. 
6 0
3 years ago
Read 2 more answers
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