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BabaBlast [244]
3 years ago
13

Need math help for this

Mathematics
1 answer:
Burka [1]3 years ago
7 0

Answer:

<h2>In the attachment</h2>

Step-by-step explanation:

The point-slope form of an equation of a line:

y-y_1=m(x-x_1)

m - slope

(x₁, y₁) - point

We have the equation:

y-5=-\dfrac{2}{3}(x+9)\\\\y-5=-\dfrac{2}{3}(x-(-9))

Therefore we have

the slope m = -2/3

and the point (-9, 5)

A slope

m=\dfrac{rise}{run}

rise = -2

run = 3

From the point (-9, 5) ⇒ 2 units down and 3 units to the right.

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Answer:

Sorry I don't remember the answer

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3 years ago
1. write it on a piece of paper
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2:-

\\ \sf\longmapsto 4\dfrac{3}{7}+6\dfrac{1}{5}

\\ \sf\longmapsto \dfrac{31}{7}+\dfrac{31}{5}

\\ \sf\longmapsto \dfrac{155+225}{35}

\\ \sf\longmapsto \dfrac{380}{35}

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3:-

\\ \sf\longmapsto 4\dfrac{2}{5}+5\dfrac{2}{6}

\\ \sf\longmapsto \dfrac{22}{5}+\dfrac{32}{6}

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4:-

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7 0
3 years ago
Read 2 more answers
What is the LCM of two numbers that have no common factors greater than one give an example
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4 0
3 years ago
If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"
seraphim [82]

Answer:

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Step-by-step explanation:

we have:

\tan(a)  +  \tan(b)  +  \tan(c)  \\  =  \tan(a)  +  \tan(b)  -  \tan(a + b)  \\  =  \tan( a)  +  \tan(b)  -  \frac{ \tan(a) +  \tan(b)  }{1 -  \tan(a)  \tan(b) }  \\  =  \frac{ ( \tan(a) +  \tan(b)  ) \tan(a) \tan(b)  }{ \tan(a) \tan(b)  - 1 } (1)

we also have:

\tan(a)  \tan(b)  \tan(c)  \\  =  -  \tan(a)  \tan(b)  \tan(a + b)  \\  =  \frac{ -(\tan( a  )   + \tan(b) ) \tan(a)  \tan(b) }{1 -  \tan(a)  \tan(b) }  \\  =  \frac{( \tan(a)  +  \tan(b)) \tan(a)   \tan(b) }{ \tan(a) \tan(b)  - 1 } (2)

from (1)(2) => proven

5 0
2 years ago
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