Which equation represents a line which is perpendicular to the line 5y-4x=-15 5y−4x=−15?

1 answer:

4 0

Y = -5/4x + b is the equation that represents a line that is perpendicular to the line 5y-4x=-15

If you want it in standard form, it is 5x+4y=-15

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A mirror has an area of 66 inches. If the mirror is 11 inches long,how wide is it?

serious [3.7K]

Answer: 6

Step-by-step explanation: with area you multiply length by width so 11 x 6 is 66

8 0

1 year ago

Please help!

Liono4ka [1.6K]

The equation
a/4 = 2/3
do cross multiplication
8 = 3a
a = 8/3

Performing the expression.
= (a - 4) / 4
= (8/3 - 4) / 4
= (8-12 / 3)/4
= -4/3 / 4
= -1 / 3

So the answer is letter b.

6 0

4 years ago

Given the point and slope, write the equation of the line

Ratling [72]

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-4}{6-1}\implies \cfrac{-5}{5}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-1(x-1) \\\\\\ y-4=-x+1\implies y=-x+5$

$\bf (\stackrel{x_1}{-12}~,~\stackrel{y_1}{14})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-14}{6-(-12)}\implies \cfrac{-15}{6+12}\implies \cfrac{-15}{18}\implies -\cfrac{5}{6}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-14=-\cfrac{5}{6}[x-(-12)] \\\\\\ y-14=-\cfrac{5}{6}(x+12)\implies y-14=-\cfrac{5}{6}x-10\implies y=-\cfrac{5}{6}x+4$