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Kipish [7]
3 years ago
5

Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely

many solutions. You can select 'always', 'never', 'a = ', or 'a ≠', then specify a value or comma-separated list of values. 2x1−6x2−4x3 = 6 −x1+ax2+4x3 = −1 2x1−5x2−2x3 = 9
Mathematics
1 answer:
navik [9.2K]3 years ago
4 0

Answer:

Never

Never

Never

Step-by-step explanation:

The equations given are

2x1−6x2−4x3 = 6 ....... (1)

−x1+ax2+4x3 = −1 ........(2)

2x1−5x2−2x3 = 9 ..........(3)

the values of a for which the system of linear equations has no solutions

Let first add equation 1 and 2. Also equation 2 and 3. This will result to

X1 + (a X2 - 6X2) - 0 = 5

And

X1 + (aX2-5X2) + 2X3 = 8

Since X2 and X3 can't be cancelled out, we conclude that the value of a is never.

a unique solution,

Let first add equation 1 and 2. Also equation 2 and 3. This will result to

X1 + (a X2 - 6X2) - 0 = 5

And

X1 + (aX2-5X2) + 2X3 = 8

The value of a = never

infinitely many solutions. 

Divide equation 1 by 2 we will get

X1 - 3X2 - 2X3 =3

Add the above equation with equation 3. This will result to

3X1 - 8X2 - 4X3 = 12

Everything ought to be the same. Since they're not.

Value of a = never.

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Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

h(t)=-16t^{2}+25t+5

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

h(t)=-16(t^{2}-\frac{25}{16}t)+5

Complete the square

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}

Rewrite as perfect squares

h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

The vertex is the point (\frac{25}{32},\frac{945}{64})

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

16(t-\frac{25}{32})^{2}=\frac{945}{64}

(t-\frac{25}{32})^{2}=\frac{945}{1,024}

square root both sides

(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}

t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}

the positive value is

t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec

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Step-by-step explanation:

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Step-by-step explanation:

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