Order the fractions, decimals, and percents from least to greatest. ( If you get it right I will mark you Brainliest.

Mathematics

1 answer:

alex41 [277]2 years ago

5 0

Step-by-step explanation:

Convert all into decimals:

0.14, 4/8=0.5,0.33

Answer:

0.14, 0.33, 0.5

0.14, 3/9, 4/8

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Question 8 Find the unit vector in the direction of (2,-3). Write your answer in component form. Do not approximate any numbers

slamgirl [31]

Answer:

The unit vector in component form is $\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$ or $\hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j$ .

Step-by-step explanation:

Let be $\vec u = (2,-3)$ , its unit vector is determined by following expression:

$\hat {u} = \frac{\vec u}{\|\vec u \|}$

Where $\|\vec u \|$ is the norm of $\vec u$ , which is found by Pythagorean Theorem:

$\|\vec u\|=\sqrt{2^{2}+(-3)^{2}}$

$\|\vec u\| = \sqrt{13}$

Then, the unit vector is:

$\hat{u} = \frac{1}{\sqrt{13}} \cdot (2,-3)$

$\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$

The unit vector in component form is $\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$ or $\hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j$ .

6 0

3 years ago

Start time 3:30 pm end time 7:00 pm what is elapsed time?

DiKsa [7]

Hello there!

Your answer is 3 hours and 30 minutes.

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3 years ago

How to solve logarithmic equations as such

Serga [27]

$\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5$

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