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Makovka662 [10]

3 years ago

12

given an example of a linear system for which (e^-t, a) is a soluotion for every constant a. sketch the direction field for this

system.

1 answer:

Harrizon [31]3 years ago

4 0

Answer:

Please see attachment

Step-by-step explanation:

Please see attachment

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Solve for x when 4x+16/12=x-4

stira [4]

Answer:

In short, Your Answer would be either 1 or 7

Step-by-step explanation:

|4x + 12| = 16

As it is in modulus function, it could be either +ve or -ve

4x+12 = 16 OR 4x - 12 = 16

4x = 16-12 OR 4x = 16 + 12

4x = 4 OR 4x = 28

x = 4/4 OR x = 28/4

8 0

2 years ago

How would I solve 5(x -2) =2x +16?

Alborosie

Answer: First, multiply 5 by x-2 to get 5x-10. We now have 5x-10=2x+16. Subtract 2x on both sides: 3x-10=16. Add 10 to both sides: 3x=26. Lastly, divide by 3x to isolate x: x=8.6666667

5 0

3 years ago

Graph a parabola whose x-intercepts are at x=-3 and x=5 and whose minimum value is y=-4

docker41 [41]

Answer:

(See explanation for further details)

Step-by-step explanation:

The standard equation of the parabola is:

$y + 4 = C \cdot (x-k)^{2}$

The formula is now expanded into a the form of a second-order polynomial:

$y + 4 = C\cdot x^{2} -2\cdot C\cdot k \cdot x +C\cdot k^{2}$

$y = C\cdot x^{2} - (2\cdot C \cdot k) \cdot x + (C\cdot k^{2}-4)$

The general equation of the second-order polynomial is:

$x = \frac{2\cdot C \cdot k \pm \sqrt{4\cdot C^{2}\cdot k^{2}-4\cdot C\cdot (C\cdot k^{2}-4)}}{2\cdot C}$

$x = k \pm \frac{\sqrt{C^{2}\cdot k^{2}-C^{2}\cdot k^{2}+4\cdot C}}{C}$

$x = k \pm 2\cdot \frac{\sqrt{C}}{C}$

$x = k \pm \frac{2}{\sqrt{C}}$

The equations to be solved are presented herein:

$-3 = k -\frac{2}{\sqrt{C}}$

$5 = k + \frac{2}{\sqrt{C}}$

Now, the solution of the system is:

$-3 +\frac{2}{\sqrt{C}} = 5 -\frac{2}{\sqrt{C}}$

$\frac{4}{\sqrt{C}} = 8$

$\sqrt{C} = \frac{1}{2}$

$C = \frac{1}{4}$

$k = 5 - \frac{2}{\sqrt{\frac{1}{4} }}$

$k = 1$

The equation of the parabola is:

$y = \frac{1}{4}\cdot (x-1)^{2} -4$

Lastly, the graphic of the function is included as attachment.

3 0

3 years ago

What is the solution to the inequality 3x - 12| ≥ 6?

natta225 [31]

<h3>Answer: x ≤ 2 or x ≥ 6</h3>

========================================

Work Shown:

|3x-12| ≥ 6

3x-12 ≥ 6 or 3x-12 ≤ -6 ..... see note below

3x ≥ 6+12 or 3x ≤ -6+12

3x ≥ 18 or 3x ≤ -6+12

3x ≥ 18 or 3x ≤ 6

x ≥ 18/3 or x ≤ 6/3

x ≥ 6 or x ≤ 2

x ≤ 2 or x ≥ 6

----

Note: if |A| ≥ B, then A ≥ B or A ≤ -B where B is some positive number.

Example: |x| ≥ 5 means either x ≥ 5 or x ≤ - 5

4 0

2 years ago

Divide 2 over 3 ÷ 1 over 6 = ____.

Ad libitum [116K]

$\frac{2}{3} \div \frac{1}{6}$
{is the same as / equal to:}
$\frac{2}{3} \times \frac{6}{1} \\ = \frac{2}{3} \times 6 \\ = \frac{2 \times 6}{3} \\ = \frac{12}{3}$
{Simplify the fraction to become:}
$= 12 \div 3 \\ = 4$

Hope this helps! :)

6 0

3 years ago