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SCORPION-xisa [38]

3 years ago

12

HELP ME PLEASE !!!!!!!!

1 answer:

german3 years ago

4 0

Answer:

the answer is the top one and the last one. Hope it helped :)

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Identify the sequence as arithmetic or geometric. f(n+1)=17 f(n), for n≥1 and f(1)=49

kaheart [24]

Answer:

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Step-by-step explanation:

6 0

3 years ago

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Find the errors and correct the following mathematical sentences.

Sergio039 [100]

Answer:

Here is another one solved.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=Given%20%5C%3A%20cotA%20%3D%20%5Csqrt%7B%5Cdfrac%7B1%7D%7B3%7D%7D" id="TexFormula1" title="Giv

ruslelena [56]

<h3>Given :</h3>

$\tt cotA = \sqrt{ \dfrac{1}{3}}$

$\tt \implies cotA = \dfrac{1}{\sqrt{3}}$

<h3>To Find :</h3>

All other trigonometric ratios, which are :

sinA
cosA
tanA
cosecA
secA

<h3>Solution :</h3>

Let's make a diagram of right angled triangle ABC.

Now, From point A,

AC = Hypotenuse

BC = Perpendicular

AB = Base

$\tt We \: are \: given, \: cotA = \dfrac{1}{\sqrt{3}}$

$\tt We \: know \: that \: cot \theta = \dfrac{base}{perpendicular}$

$\tt \implies \dfrac{base}{perpendicular} = \dfrac{1}{\sqrt{3}}$

$\tt \implies \dfrac{AB}{BC} = \dfrac{1}{\sqrt{3}}$

$\tt \implies AB = 1x \: ; \: BC = \sqrt{3}x \: (x \: is \: positive)$

Now, by Pythagoras' theorem, we have

AC² = AB² + BC²

$\tt \implies AC^{2} = (1x)^{2} + (\sqrt{3}x)^{2}$

$\tt \implies AC^{2} = 1x^{2} + 3x^{2}$

$\tt \implies AC^{2} = 4x^{2}$

$\tt \implies AC = \sqrt{4x^{2}}$

$\tt \implies AC = 2x$

Now,

$\tt sin \theta = \dfrac{perpendicular}{hypotenuse}$

$\tt \implies sinA = \dfrac{BC}{AC}$

$\tt \implies sinA = \dfrac{\sqrt{3}x}{2x}$

$\tt \implies sinA = \dfrac{\sqrt{3}}{2}$

$\Large \boxed{\tt sinA = \dfrac{\sqrt{3}}{2}}$

$\tt cos \theta = \dfrac{base}{hypotenuse}$

$\tt \implies cosA = \dfrac{AB}{AC}$

$\tt \implies cosA = \dfrac{1x}{2x}$

$\tt \implies cosA = \dfrac{1}{2}$

$\Large \boxed{\tt cosA = \dfrac{1}{2}}$

$\tt tan \theta = \dfrac{perpendicular}{base}$

$\tt \implies tanA = \dfrac{BC}{AB}$

$\tt \implies tanA = \dfrac{\sqrt{3}x}{1x}$

$\tt \implies tanA = \sqrt{3}$

$\Large \boxed{\tt tanA = \sqrt{3}}$

$\tt cosec \theta = \dfrac{hypotenuse}{perpendicular}$

$\tt \implies cosecA = \dfrac{AC}{BC}$

$\tt \implies cosecA = \dfrac{2x}{\sqrt{3}x}$

$\tt \implies cosecA = \dfrac{2}{\sqrt{3}}$

$\Large \boxed{\tt cosecA = \dfrac{2}{\sqrt{3}}}$

$\tt sec \theta = \dfrac{hypotenuse}{base}$

$\tt \implies secA = \dfrac{AC}{AB}$

$\tt \implies secA = \dfrac{2x}{1x}$

$\tt \implies secA = 2$

$\Large \boxed{\tt secA = 2}$

5 0

3 years ago

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Can you relate this lesson to your real-life situation

kati45 [8]

Answer:

This is nothing related to real life. The only thing is it helps with trigometry, programming shapes, and so on

Step-by-step explanation:

3 0

3 years ago

Help will give brainly and thanks <br> I have 5 mins

soldi70 [24.7K]

Answer:

no

Step-by-step explanation:

7 0

3 years ago

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