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3 years ago

I could probably figure this out but I have way too much homework so It'd be nice to finish this pathway.

Mathematics

1 answer:

pentagon [3]3 years ago

8 0

Yeah that's right always be a hardworking person

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A. Show that the vector v = ai + bj is perpendicular to the line ax + by = c by establishing that the slope of v is the negative

pishuonlain [190]

Answer:

Part A:

$m_1m_2=-1$

$\frac{b}{a}(\frac{-a}{b})=-1\\-1=-1$

Hence proved that Vector= ai + bj is perpendicular to the line ax + by = c.

Part B:

Slope of vector = $\frac{b}{a}$

Step-by-step explanation:

Condition for perpendicular is:

$m_1m_2=-1$

Part A:

Consider the vector v = ai + bj

x component of vector=a

y component of vector=b

Slope of vector= $m_1=\frac{y}{x}=\frac{b}{a}$

Consider the line ax + by = c:

Rearranging the equation:

ax+by=c

by=c-ax

y= $\frac{-ax}{b}+\frac{c}{b}$

According to general equation of line: $y=mx+c$

Where m is the slope

In our case the slope of above line is:

$m_2=\frac{-a}{b}$

According to the condition of perpendicular:

$m_1m_2=-1$

$\frac{b}{a}(\frac{-a}{b})=-1\\-1=-1$

Hence proved that Vector= ai + bj is perpendicular to the line ax + by = c.

Part B:

Slope of vector is also calculated above.

Since the slope of vector is negative reciprocal of the slope of the given line:

According to equation of line ax + by = c

y= $\frac{-ax}{b}+\frac{c}{b}$

According to general equation of line: $y=mx+c$

Where m is the slope

Slope of given line=m= $\frac{-a}{b}$

negative reciprocal of the slope of the given line = $\frac{b}{a}$

Slope of vector = $\frac{b}{a}$

5 0

3 years ago

Consider the first five steps of the derivation of the Quadratic Formula.

lara31 [8.8K]

Answer:

Full proof below

Step-by-step explanation:

$\displaystyle ax^2+bx+c=0\\\\ax^2+bx=-c\\\\x^2+\biggr(\frac{b}{a}\biggr)x=-\frac{c}{a}\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\biggr(\frac{b}{2a}\biggr)^2=-\frac{c}{a}+\biggr(\frac{b}{2a}\biggr)^2\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=\frac{b^2-4ac}{4a^2}\\ \\x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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Find the slope of the line passing through the points (2,5) and (0,-4).

Volgvan

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-4}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-4-5}{0-2}\implies \cfrac{-9}{-2}\implies \cfrac{9}{2}\implies 4\frac{1}{2}$

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