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4 years ago

I'm struggling with this one

Mathematics

2 answers:

mezya [45]4 years ago

8 0

Answer:

it's the third one cause its square root so x can never be less then 0 but it can go in infinitely

Ede4ka [16]4 years ago

4 0

The third 1111111111111111111111111111111111

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3 years ago

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gtnhenbr [62]

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196101

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3 years ago

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3 years ago

Read 2 more answers

In right triangle ABC, mC - 90° and AC BC. Which trigonometric ratio is cquivalent to sin b?

vaieri [72.5K]

Answer:

All trigonometric Ratios are $SinB = \frac{AC}{AB}$ , $SinA= \frac{CB}{AB}$ , $CosA= \frac{AC}{AB}$

And $Cos B = \frac{CB}{AB}$ .

Step-by-step explanation:

Given that,

A right angle triangle ΔABC, ∠C =90°.

Diagram of the given scenario shown below,

In triangle ΔABC :-

$Hypotenuse = AB\\Base = CB\\Perpendicular = AC$

So, $Sin\theta = \frac{perpendicular}{hypotenuse}$

$SinB = \frac{AC}{AB}$

Now, for ∠A the dimensions of trigonometric ratios will be changed.

Here the base for ∠A is AC , perpendicular side is CB and hypotenuse will be same for all ratios.

$SinA= \frac{CB}{AB}$

Again, $Cos\theta= \frac{base}{hypotenuse}$

Then, $CosA= \frac{AC}{AB}$

And $Cos B = \frac{CB}{AB}$ .

Hence,

All trigonometric Ratios are $SinB = \frac{AC}{AB}$ , $SinA= \frac{CB}{AB}$ , $CosA= \frac{AC}{AB}$

And $Cos B = \frac{CB}{AB}$ .

8 0

4 years ago

I'm struggling with this one ​

I'm struggling with this one