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Mathematics

2 answers:

Aleks [24]3 years ago

8 0

It's the first one you can just add or multiply the expressions to get the answer

Natasha2012 [34]3 years ago

8 0

Answer:

The answer is a

Step-by-step explanation:

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Please only answer if you know the correct answer! Will give brainliest and points!

Mandarinka [93]

Step-by-step explanation:

845/1160 = 0.728

116.5/174 = 0.669

42.25/58 = 0.728

23.4/29 = 0.806

16.9/23.2 = 0.728

25.35/34.8 = 0.728

Hence the 2nd, 4th and 5th options are correct.

7 0

2 years ago

Read 2 more answers

Evaluate : limx→ tan2-sin2x x3

GrogVix [38]

Given:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Solve:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Use l'hopital's rule:

$\begin{gathered} =\lim _{x\to0}\frac{\frac{d}{dx}(-\sin 2x+\tan 2x)}{\frac{d}{dx}(x^3)} \\ =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \end{gathered}$

Simplify:

$\begin{gathered} =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \\ =\lim _{x\to0}\frac{2(-\cos (2x)+\tan ^2(2x)+1)}{3x^2} \end{gathered}$

Apply the constant multiple rule:

$\begin{gathered} \lim _{x\to0}cf(x)=c\lim _{x\to0}f(x) \\ \text{With c=}\frac{2}{3} \\ f(x)=\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2} \end{gathered}$ $\begin{gathered} =\frac{2\lim _{x\to0}\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2}}{3} \\ =\frac{2\lim _{x\rightarrow0}\frac{(4\tan ^2(2x)+4)\tan (2x)+2\sin (2x)}{2x}}{3} \end{gathered}$

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