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solmaris [256]
2 years ago
12

Draw the unit circle with center (0,0)and plot the point P=(6,2). Observe there are TWO lines tangent to the circle passing thro

ugh the point P.
1)the tangent line on top of the circle(L1) has coordinates?

a) what is the equation for L1?

2) the tangent line below the circle(L2) has coordinates?

b)what is the equation for L2?
Mathematics
1 answer:
scZoUnD [109]2 years ago
8 0

Answer:

Step-by-step explanation:

Given that a tangent is drawn from P (6,2) to unit circle with center at the origin.

The tangent passing through (6,2) would have equaiton of the form

y-6 = m(x-2)\\y =mx-2m+6 for suitable m.

Because this line is tangent, distance of centre of circle namely origin is the radius 1.

i.e. |\frac{-2m+6}{\sqrt{1+m^2} } |=1\\(-2m+6 )^2 = 1+m^2\\3m^2-24m+35=0\\\\m=1.918, 6.082

the coordinates are the intersection of the tangent with the circle

They are (-0.162, 0.987) and (0.462,-0.887)

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C. 270 grams

Step-by-step explanation:

-This is an exponential growth function which can be expressed using the formula:

P_t=P_oe^{rt}

Where:

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Given the size at t=2 and at t=0, we substitute in the growth function to solve for r:

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1 year ago
Find the weighted average of these values
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The coordinates of the vertices of quadrilateral ABCD are A(−6, 3) , B(−1, 5) , C(3, 1) , and D(−2, −2) . Which statement correc
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Answer:

C.Quadrilateral ABCD is not a rhombus because there are no pairs of parallel sides.

Complete question:

A. Quadrilateral ABCD is not a rhombus because opposite sides are parallel but the four sides do not all have the same length.

B. Quadrilateral ABCD is a rhombus because opposite sides are parallel and all four sides have the same length.

C. Quadrilateral ABCD is not a rhombus because there are no pairs of parallel sides.

D. Quadrilateral ABCD is not a rhombus because there is only one pair of opposite sides that are parallel.

Step-by-step explanation:

Rhombus states that a parallelogram with four equal sides and sometimes one with no right angle.

Given: The coordinate of the vertices of quadrilateral ABCD are A(−6, 3) , B(−1, 5) , C(3, 1) , and D(−2, −2) .

The condition for the segment (x_{1},y_{1}), (x_{2},y_{2})  to be parallel to (x_{3},y_{3}),  (x_{4},y_{4}) is matching slopes;

\frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{y_{4}-y_{3}}{x_{4}-x_{3}} \\(y_{2}-y_{1}) \cdot (x_{4}-x_{3}) =(y_{4}-y_{3}) \cdot (x_{2}-x_{1})---->1

So, we have to check that AB || CD and AD || BC

First check AB || CD

A(−6, 3) , B(−1, 5) , C(3, 1) , and D(−2, −2)

substitute in [1],

(5-3) \cdot (-2-3) = (-2-1) \cdot (-1-(-6))2 \cdot -5 = -3 \cdot 5

-10 ≠ -15

Similarly,

check AD || BC

A(−6, 3) , D(−2, −2) , B(−1, 5) and C(3, 1)

Substitute in [1], we have

(-2-3) \cdot (3-(-1)) = (1-5) \cdot (-2-(-6))-5 \cdot 4 = -4 \cdot 4

-20 ≠ -16.

Both pairs of sides are not parallel,

therefore, Quadrilateral ABCD is not a rhombus because there are no pairs of parallel sides.

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