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Molodets [167]
3 years ago
9

Hey can someone answer this ASAP

Mathematics
1 answer:
Yakvenalex [24]3 years ago
6 0

Answer:

csc θ = \sqrt{85}/6

sec θ = \sqrt{85}/7

tan θ = 6/7

Step-by-step explanation:

The first thing to do will be to compute the length of the hypotenuse using the Pythagoras theorem;

6^2 +7^2 = hypotenuse^2

hypotenuse = \sqrt{85}

csc θ = 1/sinθ

sinθ = opposite side/hypotenuse

       = 6/\sqrt{85}

csc θ = \sqrt{85}/6

sec θ = 1/cosθ

cosθ = adjacent side/hypotenuse

sec θ = hypotenuse/adjacent side

         = \sqrt{85}/7

tan θ = Opposite side/adjacent side

         = 6/7

   

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zlopas [31]

Let

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10years after

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ATQ

\\ \tt\hookrightarrow 6x+20=2(x+20)

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3 years ago
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Passes through (6 1) and is perpendicular to the line 3x-4y=2 standard form
Snowcat [4.5K]

Answer:

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Step-by-step explanation:

Convert the Standard Form of the given line to Slope-Intercept Form:

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Then, after getting your equation, take the OPPOSITE MULTIPLICATIVE INVERSE of ¾ [perpendicular lines have OPPOSITE MULTIPLICATIVE INVERSE <em>Rate</em><em> </em><em>of</em><em> </em><em>Changes</em><em> </em>(<em>SLOPES</em>)], which is -1⅓ [or -4⁄3]. After getting your slope, you can perform the action with your given ordered pair to find your y-intercept:

1 = -1⅓[6] + b

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Now that you have your y-intercept, this is your new equation in <em>Slope-Intercept Form</em>:

y = -1⅓x + 9

We are not done just yet. The final step is to convert the Slope-Intercept Form of the line to Standard Form, and here is how it is done:

y = -1⅓x + 9

+1⅓x +1⅓x

--------------

1⅓x + y = 9 → Final equation

If you are ever in need of assistance, do not hesitate to let me know by subscribing to my channel You-Tube channel [USERNAME: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

**On my channel, I have a video that pretty much explains how to something like this. It is the longest video on there. You should be capable of finding it in a jip!

5 0
3 years ago
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kotykmax [81]
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Here, sqrt(3) is the longer leg (it's longer than 1).  
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