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just olya [345]
2 years ago
7

Which congruence criteria is met in the following scenario?

Mathematics
1 answer:
Arisa [49]2 years ago
7 0
None because the triangles have different lines
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What is the difference between adding and subtracting fractions and multiplying and dividing fractions?
77julia77 [94]
With adding and subtracting fractions, you must find a common denominator. In multiplication, you just multiply numerator by numerator; denominator by denominator. In division, you multiply the fraction by the reciprocal of the 2nd. In multiplication and division, finding a common denominator is not needed.
4 0
3 years ago
Larry the librarian and Courtney the lawyer are having a debate. Larry contends that the library is bigger than the courthouse i
professor190 [17]

Answer:

Library's Area= 5200

Courthouse' Area= 5027

<u>Larry</u> is correct.

Step-by-step explanation:

5 0
3 years ago
Pls help this is due today
Ira Lisetskai [31]

Answer: 256

Step-by-step explanation:

8x8=64

64x4=256

7 0
2 years ago
Read 2 more answers
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
Prove the theorem (AB )^T= B^T. A^T
Lisa [10]

Answer:

(AB)^T = B^T.A^T  (Proved)

Step-by-step explanation:

Given  (AB )^T= B^T. A^T;

To prove this expression, we need to apply multiplication law, power law and division law of indices respectively, as shown below.

(AB)^T = B^T.A^T\\\\Start, from \ Right \ hand \ side\\\\B^T.A^T = \frac{B^T.A^T}{A^T}.\frac{B^T.A^T}{B^T} (multiply \ through) \\\\                = \frac{A^{2T}.B^{2T}}{A^T.B^T} \\\\=\frac{(AB)^{2T}}{(AB)^T} \ \ (factor \ out \ the power)\\\\= (AB)^{2T-T}  \ (apply \ division \ law \ of \ indices; \ \frac{x^a}{x^b} = x^{a-b})\\\\= (AB)^T \ (Proved)

3 0
3 years ago
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