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AnnyKZ [126]
3 years ago
15

Suppose a system of linear equations has a 3 5 augmented matrix whose fifth column is a pivot column. is the system consistent?

why (or why not)? 4
Mathematics
1 answer:
likoan [24]3 years ago
7 0

Answer:

The system will be inconsistent.

Step-by-step explanation:

We are given that a system of linear equations has a 3×5 augmented matrix whose fifth column is a pivot column.

Then such a system is not consistent because since the augmented matrix has a pivot in fifth column it means that the new column added to the matrix A will lead to increase in the rank as that of matrix  A.

Hence the rank of A and Augment A will not remain same and hence the system will be inconsistent.

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If you roll a standard six sided dice, what is the probability you will roll a four?
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Answer:

1/6

Step-by-step explanation:

There are 6 sides on a dice. Every time you roll you you will have one of those sides. So it is a 1 in 6 chance of you getting 4, meaning that the answer to this problem is 1/6

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Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

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