Help please! find the value of y

Mathematics

1 answer:

elena-s [515]2 years ago

7 0

Answer:

Step-by-step explanation:

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1/2 (89-18) = -57<br> g=

Furkat [3]

Answer:

g= - 1/8094

Step-by-step explanation:

5 0

1 year ago

Simplify the expression

Stolb23 [73]

Answer:

A. ${x}^{3} - x - \frac{6}{2x + 3}$

Step-by-step explanation:

$\frac{2 {x}^{4} + 3 {x}^{3} - 2 {x}^{2} - 3x - 6}{2x + 3} \\ \\ = \frac{(2 {x}^{4} + 3 {x}^{3} ) - (2 {x}^{2} + 3x) - 6}{2x + 3} \\ \\ = \frac{ {x}^{3} (2 {x} + 3 ) - x(2 {x} + 3) - 6}{2x + 3} \\ \\ = \frac{ {x}^{3} (2x + 3)}{2x + 3} - \frac{x(2x + 3)}{2x + 3} - \frac{6}{2x + 3} \\ \\ \purple { \bold{= {x}^{3} - x - \frac{6}{2x + 3} }}$

8 0

2 years ago

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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