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Burka [1]
3 years ago
5

What is the slope of the line shown below?

Mathematics
1 answer:
IgorLugansk [536]3 years ago
8 0

Answer:

3/4

Step-by-step explanation:

I think

Hope this helps

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√secA+tanA/√secA-tanA × √cosecA-1/√cosecA+1=1
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Use:\\\\\sec A=\dfrac{1}{\cos A}\\\\\tan A=\dfrac{\sin A}{\cos A}\\\\\csc A=\dfrac{1}{\sin A}\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\---------------------------------\\\\\sec A+\tan A=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\dfrac{1+\sin A}{\cos A}\\\\\sec A-\tan A=\dfrac{1-\sin A}{\cos A}\\\\\csc A-1=\dfrac{1}{\sin A}-\dfrac{\sin A}{\sin A}=\dfrac{1-\sin A}{\sin A}\\\\\csc A+1=\dfrac{1+\sin A}{\sin A}


\dfrac{\sqrt{\sec A+\tan A}}{\sqrt{\sec A-\tan A}}\cdot\dfrac{\sqrt{\cos A-1}}{\sqrt{\cos A+1}}=1\\\\L_s=\sqrt{\dfrac{\sec A+\tan A}{\sec A-\tan A}\cdot\dfrac{\cos A-1}{\cos A+1}}=\sqrt{\dfrac{\frac{1+\sin A}{\cos A}}{\frac{1-\sin A}{\cos A}}\cdot\dfrac{\frac{1-\sin A}{\sin A}}{\frac{1+\sin A}{\sin A}}}\\\\=\sqrt{\dfrac{1+\sin A}{\cos A}\cdot\dfrac{\cos A}{1-\sin A}\cdot\dfrac{1-\sin A}{\sin A}\cdot\dfrac{\sin A}{1+\sin A}}\\\\\text{Everything are simplified}\\\\=\sqrt{1}=1=R_s

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Find the slope of the line. On a coordinate plane, a line goes through (0, negative 2) and (2, negative 4). a. -1 c. -2 b. 2 d.
Sedbober [7]

Answer: Choice A) Slope = -1

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

Explanation:

The two points given to us are (x1,y1) = (0,-2) and (x2,y2) = (2,-4)

Subtract the y coordinates

y2-y1 = -4-(-2) = -4+2 = -2

Subtract the x coordinates in the same order

x2-x1 = 2-0 = 2

Divide the two results to get the slope m

m = (y2-y1)/(x2-x1)

m = -2/2

m = -1

3 0
3 years ago
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