3 years ago

I need help with this question.

Mathematics

1 answer:

Kitty [74]3 years ago

4 0

i'm not sure how to solve the first one but the second is 150

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Can someone plz help me doing this :D

Mars2501 [29]

Ok so the first thing you would do is input the number where The variables are in the equation so 8^2-5(4)/4

6 0

3 years ago

Solve the system by the elimination method. 2x + y - 4 = 0 2x - y - 4 = 0 When you eliminate y, what is the resulting equation?

AveGali [126]

4x=8
This is because when you add the two expression together 2x+2x=4x the y values cancel and -4+-4=-8
4x-8=0 is equivalent to 4x=8

6 0

3 years ago

Read 2 more answers

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.