Find the slope of the line that passes through (-2, 7) and (-2, 9)

Mathematics

2 answers:

Margaret [11]3 years ago

6 0

Answer:

Step-by-step explanation:

<u><em>SLOPE FORMULA:</em></u>

$\Rightarrow \displaystyle \mathsf{\frac{Y_2-Y_1}{X_2-X_1}=\frac{RISE}{RUN} }}$

Y₂=9

Y₁=7

X₂=(-2)

X₁=(-2)

Solve.

$\displaystyle \mathsf{\frac{9-7}{(-2)-(-2)}=\frac{2}{-4}=\frac{2\div2}{-4\div2}={\mathsf{\frac{1}{ -2}=\boxed{\mathsf{-\frac{1}{2}} }}} }}$

The slope is -1/2.

igor_vitrenko [27]3 years ago

3 0

Y(2)-y(1)/x(2)-x(1)

9-7/-2–(-2)

0

It is a vertical line

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If cos x= 5/13 and sin x <0 find cos (x/2) and sin (2x)

taurus [48]

We know that

if cos x is positive
and
sin x is negative
so
the angle x belong to the IV quadrant

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we know that
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sin x=√(144/169)-------> sin x=12/13
but remember that x is on the IV quadrant
so
sin x=-12/13

Part A) <span>cos (x/2)
cos (x/2)=(+/-)</span>√[(1+cos x)/2]
cos (x/2)=(+/-)√[(1+5/13)/2]
cos (x/2)=(+/-)√[(18/13)/2]
cos (x/2)=(+/-)√[36/13]
cos (x/2)=(+/-)6/√13-------> cos (x/2)=(+/-)6√13/13
the angle (x/2) belong to the II quadrant
so
cos (x/2)=-6√√13/13

the answer Part A) is
cos (x/2)=-6√√13/13

Part B) sin (2x)
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the answer Part B) is
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3 years ago

Eliminate the parameter. x = 5t, y = t + 8

uysha [10]

$\bf \begin{cases}
x=5t\implies \frac{x}{5}=\boxed{t}\\
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----------\\
y=\boxed{\frac{x}{5}}+8
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