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2 years ago

Find the value of x pleasee

Mathematics

2 answers:

Citrus2011 [14]2 years ago

5 0

Answer:

x = 22

Step-by-step explanation:

For the lines to be parallel , then

(6x - 44) and 4x will be congruent ( alternate angles ) , that is

6x - 44 = 4x ( subtract 4x from both sides )

2x - 44 = 0 ( add 44 to both sides )

2x = 44 ( divide both sides by 2 )

x = 22

Charra [1.4K]2 years ago

4 0

Answer:

$4x \degree = (6x - 44)\degree \\ \{alternate \: angles \} \\ 4x = 6x - 44 \\ 6x - 4x = 44 \\ 2x = 44 \\ { \underline{ \underline{x = 22}}}$

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(-3,7)(9,5)

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2 years ago

Derivative of tan(2x+3) using first principle

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The derivative is given by the limit

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You have

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$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

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$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
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$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

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pav-90 [236]

Answer:

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