What is the slope of the line containing the points (5,6) and (9,13)

2 answers:

4 0

Hello! You find the slope of the line by doing the formula y2 - y1 / x2 - x1. That means you subtract the first y-coordinate from the second y-coordinate and the first x-coordinate from the second x-coordinate. Plug in the values to get 13 - 6 / 9 - 5. Subtract the numbers to get 7/4. There. The slope of the line is 7/4.

Nataly_w [17]2 years ago

3 0

You would subtract the y intercepts then the x intercepts. Subtract 13-6 then divide it by 9-6. That gets you a slope of 7 over 3

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What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Ipatiy [6.2K]

Answer:

$x = \frac{ 5 \ + \ \sqrt{31}}{2} \ , \ x = \frac{ 5 \ - \ \sqrt{31}}{2}$

Step-by-step explanation:

$2x^2 - 10x - 3 = 0 \\\\a = 2 \ , b = - 10 \ , \ c = - 3 \\\\x = \frac{-b^2\ \pm \ \sqrt{b^2 - 4ac}}{2a}\\\\x = \frac{10 \ \pm \sqrt{(-10)^2 - ( 4 \times 2 \times -3)} }{2 \times 2}\\\\x = \frac{10 \ \pm \sqrt{(100 - ( -24 )} }{4}\\\\x = \frac{10 \ \pm \sqrt{(100 + 24 } }{4}\\\\x = \frac{ 10 \ \pm \sqrt{124}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{4 \times 31}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{2^2 \times 31}}{4}\\\\x = \frac{ 10 \ \pm2 \sqrt{31}}{4}\\\\x = \frac{ 5 \ \pm\sqrt{31}}{2}\\\\$