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statuscvo [17]
3 years ago
13

Help please and show work

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
7 0

Answer:

Est en iangls ponel en  PEAIO LPR FAOVO qubese  est ameiradAS

Step-by-step explanation:

l ueq hay quoehcer  e trasdcuilroxd os ry no esrntendo nadaia

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If mZDCB = 140°, then mZACD = [?]° !!!PLS HELP I WILL MARK BRAINLIEST !!!
Readme [11.4K]

Answer:

40°

Step-by-step explanation:

  • m\angle DCB +m\angle ACD = 180\degree (Linear pair angles)

  • \implies 140\degree +m\angle ACD = 180\degree

  • \implies m\angle ACD = 180\degree-140\degree

  • \implies m\angle ACD = 40\degree
5 0
2 years ago
-2x + 3y = 18. x = -y + 1
Bezzdna [24]

Answer:

D. X=-3, y=4

Step-by-step explanation:

-2x+3y=18 equation 1

x=-y+1 equation 2

-2(-y+1)+3y=18 replace x value of equation 2 into equation 1

2y-2+3y=18

5y=20

y=4

solve for x by replacing y value into either equation.

x=-y+1

x=-4+1

x=-3

8 0
3 years ago
10 1/4 divided by 151 3/16
Svetlanka [38]
10 1/4 / 151 3/16 = 4/59

Have a great day :)
8 0
4 years ago
Read 2 more answers
Prove or give a counterexample:
ozzi

Answer:

See proof below.

Step-by-step explanation:

True

For this case we need to use the following theorem "If v_1, v_2,....v_k are eigenvectors of an nxn matrix A and the associated eigenvalues \lambda_1, \lambda_2,...,\lambda_k are distinct, then v_i's are linearly independent". Now we can proof the statement like this:

Proof

Let A a nxn matrix and we can assume that A has n distinct real eingenvalues let's say \lambda_1, \lambda_2, ....,\lambda_n

From definition of eigenvector for each one \lambda_i needs to have associated an eigenvector v_i for 1 \leq i \leq n

And using the theorem from before , the n eigenvectors v_1,....,v_n are linearly independent since the \lambda_i 1\leq i \leq n are distinct so then we ensure that A is diagonalizable.

4 0
3 years ago
Yes I’m posting every single question :)
Rudiy27

Answer:

first option

Step-by-step explanation:

i just looked with my eyes

6 0
4 years ago
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