What is the answer? X=

Mathematics

1 answer:

Minchanka [31]3 years ago

7 0

Answer:

Step-by-step explanation:

$\because \triangle ACB \cong \triangle DCE... (given) \\\therefore \angle A \cong \angle D... (c.a.c.t)\\\therefore m\angle A = m\angle D\\\therefore 50\degree = 10x\\\\\therefore x = \frac{50\degree}{10}\\\\\huge \orange {\boxed {\therefore x = 5\degree}}$

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Evaluate the expression for x = 10 . 3x + 4(x - 8) - 14

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Answer:

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Vadim26 [7]

Answer:

See Below.

Step-by-step explanation:

We are given the equation:

$\displaystyle y^2 = 1 + \sin x$

And we want to prove that:

$\displaystyle 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right) ^2 + y^2 = 1$

Find the first derivative by taking the derivative of both sides with respect to <em>x: </em>

<em /> $\displaystyle 2y \frac{dy}{dx} = \cos x$ <em />

Divide both sides by 2<em>y: </em>

<h3><em /> $\displaystyle \frac{dy}{dx} = \frac{\cos x}{2y}$ <em /></h3>

<em />

Find the second derivative using the quotient rule:

$\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &= \frac{(\cos x)'(2y) - (\cos x)(2y)'}{(2y)^2}\\ \\ &= \frac{-2y\sin x-2\cos x \dfrac{dy}{dx}}{4y^2} \\ \\ &= -\frac{y\sin x + \cos x\left(\dfrac{\cos x}{2y}\right)}{2y^2} \\ \\ &= -\frac{2y^2\sin x+\cos ^2 x}{4y^3}\end{aligned}$