What is the value of x in the equation

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

7 0

The answer to that is 2

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Find the value of the trigonometric ratio, tan X.

MariettaO [177]

Answer:

A. 24/7

<u>What are sine, cosine, and tangent?</u>

SOH - CAH - TOA

sine = opposite/hypotenuse

cosine = adjacent/hypotenuse

tangent = opposite/adjacent (that is not hypotenuse)

Step-by-step explanation:

tanX = ZY/XY = 48/14 = 24/7

5 0

3 years ago

Read 2 more answers

A) Use the limit definition of derivatives to find f’(x)

Ann [662]

$\text{Given that,}\\\\f(x) = \dfrac{ 1}{3x-2}\\\\\text{First principle of derivatives,}\\\\f'(x) = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{ h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3(x+h) - 2} - \tfrac 1{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3x+3h -2} - \tfrac{1}{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{3x-2-3x-3h+2}{(3x+3h-2)(3x-2)}}{h}\\\\\\$

$~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{-3h}{(3x+3h-2)(3x-2)}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{-3h}{h(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \lim \limits_{h \to 0} \dfrac{1}{(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \cdot \dfrac{1}{(3x+0-2)(3x-2)}\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)(3x-2)}\\\\\\~~~~~~~=-\dfrac{3}{(3x-2)^2}$

$\text{Given that,}~\\\\f(x) = \dfrac{1}{3x-2}\\\\\textbf{Power rule:}\\\\\dfrac{d}{dx}(x^n) = nx^{n-1}\\\\\textbf{Chain rule:}\\\\\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\\\\\text{Now,}\\\\f'(x) = \dfrac{d}{dx} f(x)\\\\\\~~~~~~~~=\dfrac{d}{dx} \left( \dfrac 1{3x-2} \right)\\\\\\~~~~~~~~=\dfrac{d}{dx} (3x-2)^{-1}\\\\\\~~~~~~~~=-(3x-2)^{-1-1} \cdot \dfrac{d}{dx}(3x-2)\\\\\\~~~~~~~~=-(3x-2)^{-2} \cdot 3\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)^2}$