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Firdavs [7]
3 years ago
13

Please help with this question!!

Mathematics
1 answer:
xxMikexx [17]3 years ago
8 0

\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2\\\\l\perp k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\\text{We have the points J(-24, -4) and K(-4, 6)}.\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{substitute:}\\\\m_1=\dfrac{6-(-4)}{-4-(-24)}=\dfrac{10}{20}=\dfrac{1}{2}\\\\\text{therefore}\ m_2=-\dfrac{1}{\frac{1}{2}}=-2\\\\\text{The formula of a midpoint:}\\\\\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)\\\\\text{substitute the coordinates of the points J and K:}

x=\dfrac{-24+(-4)}{2}=\dfrac{-28}{2}=-14\\\\y=\dfrac{-4+6}{2}=\dfrac{2}{2}=1\\\\\text{midpoint}\ (-14,\ 1)\\\\\text{The point-slope form:}\\\\y-y_1=m(x-x_1)\\\\\text{substitute}\ m=-2,\ x_1=-14\ \text{and}\ y_1=1:\\\\y-1=-2(x-(-14))\\\\\boxed{y-1=-2(x+14)}

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Pavel [41]

\boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}

<h2>Explanation:</h2>

Using the quadratic formula:

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ Here: \\ \\ f(x) = x^2 + 5x + 5 \\ \\ \\ So: \\ \\ a=1 \\ \\ b=5 \\ \\ c=5 \\ \\ \\ x=\frac{-5 \pm \sqrt{5^2-4(1)(5)}}{2(1)} \\ \\ x=\frac{-5 \pm \sqrt{25-20}}{2} \\ \\ x=\frac{-5 \pm \sqrt{5}}{2} \\ \\ \\ Two \ solutions: \\ \\ \boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}

<h2>Learn more:</h2>

Quadratic functions: brainly.com/question/12164750

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The points (3,-2) and (2, 1) are both solutions to which inequality?
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(3, -2) and (2, 1) are solutions to the inequality y <= 3/2x - 1

<h3>How to determine the inequality?</h3>

The complete question is added as an attachment

The points are given as:

(3, -2) and (2, 1)

Next, we test the points on each inequality in the list of options.

So, we have:

<u>Option 1</u>

y > 1/2x + 2

Substitute (3, -2) and (2, 1) for x and y

-2 > 1/2 * 3 + 2 ⇒ -2 > 3.5 -- false

2 > 1/2 * 1 + 2 ⇒ 2 > 2.5 -- false

Hence, (3, -2) and (2, 1) are not solutions to the inequality y > 1/2x + 2

<u>Option 2</u>

y <= 3/2x - 1

Substitute (3, -2) and (2, 1) for x and y

-2 <= 3/2 * 3 - 1 ⇒ -2 <= 3.5 -- true

1 <= 3/2 * 2 - 1 ⇒ 1 < 2 -- true

Hence, (3, -2) and (2, 1) are solutions to the inequality y <= 3/2x - 1

<u>Option 3</u>

y >= 4x - 2

Substitute (3, -2) and (2, 1) for x and y

-2 >= 4 * 3 - 2 ⇒ -2 >= 10 -- false

2 <= 4 * 2 - 2 ⇒ 2 <= 6 -- false

Hence, (3, -2) and (2, 1) are not solutions to the inequality y >= 4x - 2

<u>Option 4</u>

y < -2x + 1

Substitute (3, -2) and (2, 1) for x and y

-2 < -2 * 3 + 1 ⇒ -2 < -5 -- false

2 < -2 * 2 + 1 ⇒ 2 < -3 -- false

Hence, (3, -2) and (2, 1) are not solutions to the inequality y < -2x + 1

Read more about inequality at

brainly.com/question/24372553

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