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Advocard [28]
3 years ago
15

Alejandro made an error in the steps below when determining the equation of the line that is perpendicular to the line 4x – 3y =

–8 and passes through the point (3, –2).
Mathematics
2 answers:
vazorg [7]3 years ago
4 0
The slope of the line given its equation is calculated through, m = -A / B. The slope of the given line is 4/3. The line perpendicular to it has the slope of -3/4. The slope-point form of the equation is, 
                                       y - y1 = m(x - x1)
where m is the slope and x1 and y1 the abscissa and ordinate of the point, respectively. 
Substituting the values above, 
                                       y --2 = (-3/4)(x - 3)

Simplifying the equation gives 3x + 4y = 1.
gtnhenbr [62]3 years ago
4 0

The generic equation of the line is:

y-yo = m (x-xo)

Where,

(xo, yo): point where the line passes

m: slope of the line

We have the following equation:

4x - 3y = -8

Rewriting we have:

3y = 4x + 8

y = (\frac{4}{3}) x + \frac{8}{3}

Since the lines are perpendicular, then the slope of the line is the inverse reciprocal.

We have then:

m =-\frac{3}{4}

The point where the line passes is:

(xo, yo) = (3, -2)

Substituting values we have:

y + 2 = -\frac{3}{4} (x-3)

Answer:

the line that is perpendicular is:

y + 2= -\frac{3}{4}(x-3)

Note: compare with Alejandro's steps, in order to find the error.

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Read 2 more answers
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keeping in mind that 21 months is more than a year, since there are 12 months in a year,  then 21 months is really 21/12 years.


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\bf ~~~~~~ \stackrel{\textit{account B}}{\textit{Simple Interest Earned}} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &28\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ t=years\to \frac{21}{12}\dotfill &\frac{7}{4} \end{cases} \\\\\\ 28=P(0.049)\left( \frac{7}{4} \right)\implies \cfrac{28}{(0.049)\left( \frac{7}{4} \right)}=P\implies \boxed{326.53\approx P}


so, clearly, you can see who's greater.

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4 0
3 years ago
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