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Strike441 [17]
2 years ago
7

What is the equation of the line that passes through the point(-5,-4) and has a slope of -3/5?

Mathematics
1 answer:
Elodia [21]2 years ago
6 0

Answer:

y= mx+c

-4= -3/5(-5) + c

-4= -3/5 *-5+ c

-4=3+c

-4-3=c

-7=c

therefore the equation of the line is y=-3/5x-7

Step-by-step explanation:

the equation of a line is

y=mx +c

where y is the y coordinate

where m is the slope

where x is the x coordinate

where c is the y- intercept

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Find the area of the image below
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The area of a parallelogram:

A_P=5\cdot3=15

The area of a triangle:

A_T=\dfrac{1}{2}\cdot5\cdot5=12.5

The area of the polygon:

[texA=A_P+A_T\\\\A=15+12.5=27.5\ u^2[/tex]

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4 years ago
I need help proving this ASAP
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Answer:

See explanation

Step-by-step explanation:

We want to show that:

\tan(x +  \frac{3\pi}{2} )  =  -   \cot \: x

One way is to use the basic double angle formula:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ \sin(x)  \cos( \frac{3\pi}{2} )  +   \cos(x)  \sin( \frac{3\pi}{2}) }{\cos(x)  \cos( \frac{3\pi}{2} )   -    \sin(x)  \sin( \frac{3\pi}{2}) }

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We simplify further to get:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ 0  -   \cos(x) }{0 +    \sin(x) }

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\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{- \cos(x) }{ \sin(x) }

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\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  -  \cot(x)

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