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3 years ago

How would you write the following expression as a sum or difference? `"log"(root(3)(2 - x)/(3x))`

Mathematics

1 answer:

pashok25 [27]3 years ago

7 0

<h3>Further explanation</h3>

Let's recall following formula about Exponents and Surds:

$\boxed { \sqrt { x } = x ^ { \frac{1}{2} } }$

$\boxed { (a ^ b) ^ c = a ^ { b . c } }$

$\boxed {a ^ b \div a ^ c = a ^ { b - c } }$

$\boxed {\log a + \log b = \log (a \times b) }$

$\boxed {\log a - \log b = \log (a \div b) }$

<em>Let us tackle the problem!</em>

$\texttt{ }$

$\log \frac{\sqrt{3}(2-x)}{(3x)} = \log \sqrt{3} + \log (2-x) - \log (3x)$

$\log \frac{\sqrt{3}(2-x)}{(3x)} = \log 3^{1/2} + \log (2-x) - (\log 3 + \log x)$

$\log \frac{\sqrt{3}(2-x)}{(3x)} = \frac{1}{2} \log 3 + \log (2-x) - \log 3 - \log x$

$\log \frac{\sqrt{3}(2-x)}{(3x)} = \log (2-x) - \frac{1}{2}\log 3 - \log x$

$\texttt{ }$

<h3>Learn more</h3>

Coefficient of A Square Root : brainly.com/question/11337634
The Order of Operations : brainly.com/question/10821615
Write 100,000 Using Exponents : brainly.com/question/2032116

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Exponents and Surds

Keywords: Power , Multiplication , Division , Exponent , Surd , Negative , Postive , Value , Equivalent , Perfect , Square , Factor.

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For this case we propose a rule of three according to the data of the statement. To do this, we convert the mixed number into an improper fraction:

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4 ----------> $\frac {7} {4}$

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Where the variable "x" represents the amount of chopped of vegetables needed for 10 people.

$x = \frac {10 * \frac {7} {4}} {4}\\x = \frac {\frac {70} {4}} {4}\\x = \frac {70} {16}\\x = \frac {35} {8}$

This is equivalent to $4 \frac {3} {8}$ cups.

Answer:

$4 \frac {3} {8}$ cups

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Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

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NemiM [27]

Answer:

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Step-by-step explanation:

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