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Zinaida [17]
3 years ago
15

Find the equation of the line perpendicular to 3x-7y=42 which passes through the point (-3,-8)

Mathematics
1 answer:
harkovskaia [24]3 years ago
4 0
For perpendicular lines, m1m2 = -1 or m2 = -1/m1; where m1 and m2 are the slopes of the lines.
Here line 1 is 3x - 7y = 42
7y = 3x - 42
y = 3/7 x - 6; Hence m1 = 3/7
m2 = -1/(3/7) = -7/3
Required equation y - y1 = m2(x - x1)
y - (-8) = -7/3(x - (-3))
y + 8 = -7/3(x + 3)
y + 8 = -7/3 x - 7
y = -7/3 x - 7 - 8
y = -7/3 x - 15
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Exact Form:

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Decimal Form:

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Step-by-step explanation:

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First, substitute the actual values of the points into the distance formula.

Then, just simplify

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1) y= - 2x² + 8x. It's a parabola open downward (a<0)  
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To find the intersections between 1) & 2), let 1) = 2)
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3 0
3 years ago
I need help with these questions
inysia [295]

Answer:

28. m<A=20°, m<B=70°

32. m<A=103°, m<B=77°

Step-by-step explanation:

complementary angle=a+b=90°

supplementary angle=a+b=180°

28. A+B=90°

      5x+17x+2=90°

      22x+2=90°

      22x=90-2

      22x=88

        22x/22=88/22

      x=88/22=<u>4</u>

m<A=5x=5*4=<u>20°</u>

m<B=17x+2=17*4+2=68+2=<u>70°</u>

         <u>Check</u>

A+B=90°

20+70=90°

<u>90°=90°</u>

32. A+B=180°

     x+11+x-15=180°

     2x-4=180°

     2x=180+4

     2x=184

     2x/2=184/2

     x=184/2

     x=<u>92</u>

m<A=x+11=92+11=<u>103°</u>

m<B=x-15=92-15=<u>77°</u>

     <u>Check</u>

A+B=180°

103+77=180°

<u>180°=180°</u>

6 0
3 years ago
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