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katen-ka-za [31]
3 years ago
14

Which quadratic equation has the roots 0 and 3

Mathematics
1 answer:
Dima020 [189]3 years ago
4 0
Umm do we get like     A, B C D answers to help you with?
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Please help ASAP! The terminal side of an angle θ in standard position passes through the point (2,-5). calculate the values of
Troyanec [42]

Answer:

sin(x)=\frac{opp}{hyp}=\frac{5}{\sqrt{21}}=\frac{5\sqrt{21}}{21}\\cos(x)=\frac{adj}{hyp}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}\\tan(x)=\frac{opp}{adj}=\frac{5}{2}\\csc(x)=\frac{hyp}{opp}=\frac{\sqrt{21}}{5}\\sec(x)=\frac{hyp}{adj}=\frac{\sqrt{21}}{2}\\cot(x)=\frac{adj}{opp}=\frac{2}{5}

Step-by-step explanation:

Start by drawing out the triangle on the graph:

(See picture)

(By the drawing you can judge me that I am a really bad artist)

Theta and the side lengths are labeled, the hypotenuse has a length of √21 by Pythagora's theorem. Now, time to put everything to the trigonometric functions:

sin(x)=\frac{opp}{hyp}=\frac{5}{\sqrt{21}}=\frac{5\sqrt{21}}{21}\\cos(x)=\frac{adj}{hyp}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}\\tan(x)=\frac{opp}{adj}=\frac{5}{2}\\csc(x)=\frac{hyp}{opp}=\frac{\sqrt{21}}{5}\\sec(x)=\frac{hyp}{adj}=\frac{\sqrt{21}}{2}\\cot(x)=\frac{adj}{opp}=\frac{2}{5}

(Suppose that x is theta as I can't type it)

By some basic understanding that's all that I can do.

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Suppose a line has slope 4 and passes through the point (-2,5). Which other point must also be on the graph?
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Slope of 4 means you add 4 to the y and 1 to the x so the answer is (-1,9)
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