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tatiyna
3 years ago
6

-x < - 29 solve for x

Mathematics
1 answer:
AleksAgata [21]3 years ago
4 0

Answer:

x<29

Step-by-step explanation:

divide both sides by -1 since x cannot be negative

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What is 5 1/4 as a single fraction greater than 1
Sauron [17]

Answer:

21/4

Step-by-step explanation:

(5 * 4) + 1

20 + 1

21/4

7 0
3 years ago
PLEASE DONT!!Just give me the answer please explain cause I can’t learn like that NOT TRYING TO BE RUDE
icang [17]
There are approximately 4 weeks per month.

Mr. Gleason will take home 1100 x 4 weeks = $4,400 per month.

The total budget - his take home pay is the amount his wife will need to make:

5,600 - 4,400 = $1,200

Now divide the total amount she needs to make by how much she brings home per hour:

1,200 / 10 = 120

 She needs to work 120 hours per month.

The answer is D.

8 0
4 years ago
Let V be a vector space of dimension 4. Determine if each statement is true or false. (a) Any set of 5 vectors in V must be line
son4ous [18]

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
3 years ago
How to solve x=2y and 6y-3x=18 by using subsitution
BARSIC [14]

Answer:

This system of equations has no solutions

Step-by-step explanation:

in the linear equations, if the variable is disappeared, and

  • The two sides of the equation are equal, then the equation has infinitely many solutions
  • The two sides of the equation are not equal, then the equation has no solutions

Let us solve the question

∵ x = 2y ⇒ (1)

∵ 6y - 3x = 18 ⇒ (2)

→ Substitute x in equation (2) by the value of x in equation (1)

∵ 6y - 3(2y) = 18

∴ 6y - 6y = 18

∴ 0 = 18

→ 0 can not equal 18

∴ Left hand side ≠ Right hand side

→ From the 2nd note above

∴ This system of equations has no solutions

3 0
3 years ago
HELP PLZ!!! QUICK!!! I BEG YOU!!! I'll GIVE BRAINLYEST!! 10 POINTS!!
Harlamova29_29 [7]

answer: C

good luck!!!          

7 0
3 years ago
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