3 years ago

(-5, 4) and (4,1) What is the distance between the two points rounded to the nearest tenth

Mathematics

1 answer:

mr Goodwill [35]3 years ago

6 0

Answer:

AB=9.49

Step-by-step explanation:

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Please solve this and only answer with image

horsena [70]

$\cot^4 A - \cot^2 A = 1\\\\\cot^4 A = 1 + \cot^2 A\\\\\frac{\cos^4 A}{\sin^4 A} = 1 + \frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A}{\sin^2 A}+\frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A+\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{1}{\sin^2 A}\\\\\cos^4 A = \frac{\sin^4 A}{\sin^2 A}\\\\\cos^4 A = \sin^2 A\\\\$

This then means,

$\cos^4 A + \cos^2 A = 1\\\\\sin^2 A + \cos^2 A = 1\\\\$

which is the pythagorean trig identity. This concludes the proof.

Therefore, if $\cot^4 A - \cot^2 A = 1$ , then $\cos^4 A + \cos^2 A = 1$

5 0

2 years ago

-200p^2+12028p=180096

bekas [8.4K]

Answer:

Step-by-step explanation:

200 p²-12028p+180096=0

50 p-3007 p +45024=0

$p=\frac{3007 \pm\sqrt{(-3007)^2-4*50*45024} }{2*50} \\=\frac{3007 \pm \sqrt{9042049-9004800} }{100} \\=\frac{3007 \pm \sqrt{37249} }{100} \\=\frac{ 3007 \pm193}{100} \\p=32,28.14$