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tino4ka555 [31]
3 years ago
9

Let e be the line through A (-4,5) and B (3. t) that is perpendicular to the line through

Mathematics
1 answer:
grigory [225]3 years ago
8 0

Answer:

Step-by-step explanation:

product of slopes of perpendicular lines=-1

(t-5)/(3+4)×(2-3)/(-4-1)=-1

(t-5)/7×(-1/-5)=-1

(t-5)/35=-1

t-5=-1×35=-35

t=-35+5

t=-30

2.

slopes of parallel lines are equal.

(-2+3)/(t-4)=(-1-4)/(4+2)

1/(t-4)=-5/6

t-4=-6/5

t=4-6/5=(20-6)/5=14/5

3.

x>0,y<0

so P lies in4th quadrant.

except cos and sec all are negative.

so only cos and sec are positive.

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Through: (-3,2) and (0, -3)
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Answer:

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(-3-2)/(0+3) = -5/3 <--- slope

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3. Put both of those values into the new equation

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If x=(10-3i) and y=(3-10i), then xy=? and x/y=?
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xy=(10-3i)(3-10i)

To compute x times y must use foil on the right hand side.

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------------------------------------Add like terms:

-109i

----------------

\frac{x}{y}

\frac{10-3i}{3-10i}

Multiply top and bottom by bottom's conjugate:

\frac{(10-3i)(3+10i)}{(3-10i)(3+10i)}

Foil the top and just do first and last of Foil for the bottom since the bottom contains multiplying conjugates:

\frac{30+100i-9i-30i^2}{9-100i^2}

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Answer:

x = 2

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x^2 - 2x  =  x^2 + 2x - 8

Subtract 2x from both sides.

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Subtract x^2 from both sides.

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