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amid [387]
3 years ago
12

Determine the slope between the points (-1, -5) and (5, 4)

Mathematics
1 answer:
Basile [38]3 years ago
4 0

Answer:

Slope = 9/6

Step-by-step explanation:

Step 1: We have formula to find the slope if we are given two point.

<h3>slope = \frac{(y2 - y1)}{(x2 - x1)}</h3>

Given points (-1, -5) and (5, 4)

Here x1 = -1, y1 = -5 and x2=5 and y2 = 4

Step 2: Plug in those values into formula and simplify.

Slope = (4 - (-5)) / (5 - (-1))

= (4 + 5)/(5 + 1)                          -(-5)= 5 and -(-1) = 1

Slope (m) = 9/6

Thank you.

Hope you will understand this.

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Find the six trigonometric function values of the angle θ in standard position, if the terminal side of θ is defined by x + 2y =
Black_prince [1.1K]

Answer:

\sin \theta  = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta  = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta  = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta  = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}

Step-by-step explanation:

First, we need to draw the terminal position of the given angle. To do so, we need to find a point that lies on the straight line x + 2y= 0, x\geq 0

If we choose x = 2 (we can do so because of the condition x \geq 0, which means that any positive value is suitable for x), then we have

2 +2y = 0\implies 2 = -2y \implies y = -1

Therefore, the terminal side of the angle \theta  is passing through the origin and the point  (2,-1) and now we can draw it.

The angle  \theta  is presented below.

The distance of the point  (2,-1) from the origin equals

r = \sqrt{2^2 + (-1)^2} = \sqrt{5}

Now, we can determine the values of the six trigonometric function, by using their definitions.

\sin \theta  = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta  = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta  = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta  = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}

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