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Elis [28]
3 years ago
10

Add the equations. What value belongs in the green box? 3x + 2y = 9 -x - 2y = 4 [?] + [?] = [?] A. 4x B. 2x C. -3x D. -2x​

Mathematics
1 answer:
Naddika [18.5K]3 years ago
6 0

Consider the first blank is the green box.

Given:

The equations are

3x+2y=9

-x-2y=4

To find:

The value belongs in the green box

Solution:

We have,

3x+2y=9        ...(i)

-x-2y=4        ...(ii)

Adding (i) and (ii), we get

3x+2y-x-2y=9+4

(3x-x)+(2y-2y)=13

2x+0=13

So, the value for the first black is 2x. It means the value belongs in the green box is 2x.

The values of three blanks are 2x, 0 and 13 respectively.

Therefore, the correct option is B.

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Answer:

The smaller angle is 35 and the larger one is 145.

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Step-by-step explanation:

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Answer:

a) True

b) False

c) False

d) False

e) True

Step-by-step explanation:

a) Each basis of V has four vectors. Then any set of 5 vectors must be linear dependent (LD).

b) Suppose that \{v_1,v_2,v_3,v_4\} is a basis of V. Considere the set A=\{v_1,\lambda_1v_1,\lambda_2v_1,v_2,v_3\} where \lambda_1, \lambda_2 are scalars. The set has 5 vectors but V\neq span(A) because v_4 is not belong to A and v_4  is linear independent of v_1

c)  Suppose that \{v_1,v_2,v_3,v_4\} is a basis of V. Considere the set A=\{v_1,\lambda_1v_1,\lambda_2v_1,\lambda_3v_1\} where \lambda_1, \lambda_2,\lambda_3 are scalars. A has four nonzero vectors but isn't a basis because is a LD set.

d)  Suppose that \{v_1,v_2,v_3,v_4\} is a basis of V. Considere the set A=\{v_1,\lambda_1v_1,\lambda_2v_1\} where \lambda_1, \lambda_2 are scalars. A has 3 nonzero vectors but isn't a basis because is a LD set.

e)  Since any basis of V must have 4 elements, then a set of three vectors cannot generate V.

4 0
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An arxhitext built a road 4 1/3 miles long. the next road he built was 7 2/4 miles long. What is the combined lenth of the two r
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5 0
3 years ago
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