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3 years ago

Find the equation of the line that is perpendicular to the line y = (-1/3)x -1 and passes through the point (1, 5)?

Mathematics

1 answer:

Anit [1.1K]3 years ago

4 0

bearing in mind that perpendicular lines have negative reciprocal slopes, so

$\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{1}\implies 3}}$

so we're really looking for a line whose slope is 3 and runs through (1,5)

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{5})~\hspace{10em} slope = m\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-5=3(x-1) \\\\\\ y-5=3x-3\implies y=3x+2$

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What is the least common denominator for the fractions1/6and 3/4 ?

melomori [17]

A. 12; Least common denominator is the least multiple that the denominator have together. So you multiply the denominators until you get the first number they both have in common.
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3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

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3 years ago

2*2<br><img src="https://tex.z-dn.net/?f=%20%5Ccos%28300%29%20" id="TexFormula1" title=" \cos(300) " alt=" \cos(300) " align="ab

ankoles [38]

Step-by-step explanation:

$\cos(300 \degree) \\ \\ = \cos(360 \degree - 60\degree) \\ \\ = \cos( 60\degree) \\ \\ = \frac{1}{2} \\ \\ \huge \orange{ \boxed{ \therefore \: \cos(300 \degree) = \frac{1}{2} }}$

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3 years ago

Which expression is equivalent to 5 square root 11^3

AlladinOne [14]

Answer:

11^3/5

Step-by-step explanation:

The exponent of 3 becomes the numerator while the 5 in the root becomes the denominator which creates the fraction 3/5. So the final answer is 11^3/5

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3 years ago