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4 years ago

Solve −2x2 − 16x − 44 = 0.

Mathematics

2 answers:

Mashutka [201]4 years ago

5 0

<u>Answer</u>

x = -4 plus or minus "i" times the square root of 6

<u>Explanation</u>

−2x² − 16x − 44 = 0.

−2x² − 16x = 44

Dividing by -2

x² + 8x = -22

x² + 8x + 4² = -22 + 4²

(x + 4)² = -22 + 16

(x + 4) ² = -6

(x + 4) = √-6

Remember √(-1) = i

(x + 4) = i√6 or (x + 4) = -i√6

x = -4 + i√6 x = -4 - i√6

Mazyrski [523]4 years ago

3 0

I hope this helps you

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Answer:

The multiples of a whole number are found by taking the product of any counting number and that whole number. For example, to find the multiples of 3, multiply 3 by 1, 3 by 2, 3 by 3, and so on. To find the multiples of 5, multiply 5 by 1, 5 by 2, 5 by 3, and so on.

Step-by-step explanation:

5 × 1 = 5

5 × 2 = 10

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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4 years ago

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Find the slope of each line and then determine if the lines are parallel, perpendicular or neither. If a value is not an integer

Aleonysh [2.5K]

As given by the question

There are given that the point of two-line

$\begin{gathered} (1,\text{ 7) and (5, 5)} \\ (-1,\text{ -3) and (1, 1)} \end{gathered}$

Now,

From the condition of a parallel and perpendicular line

If the slopes are equal then the lines are parallel

If the slopes are negative reciprocal then the lines are perpendicular

If the slopes are neither of the above are true then lines are neither

Then,

First, find the slope of both of line

So,

For first-line, from the formula of slope

$\begin{gathered} m_1=\frac{y_2-y_1}{x_2-x_1} \\ m_1=\frac{5_{}-7_{}}{5_{}-1_{}} \\ m_1=-\frac{2}{4} \\ m_1=-\frac{1}{2} \end{gathered}$

Now,

For second-line,

$\begin{gathered} m_1=\frac{y_2-y_1}{x_2-x_1} \\ m_1=\frac{1_{}-(-3)_{}}{1-(-1)_{}} \\ m_1=\frac{4}{2} \\ m_1=2 \end{gathered}$

The given result of the slope is negative reciprocal because

$\begin{gathered} -\frac{1}{2}=-(-\frac{2}{1}) \\ -\frac{1}{2}=2 \end{gathered}$

Hence, the slope of line1 is -1/2, and slope of line2 is 2 and the lines are perpendicular.

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