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kogti [31]
10 months ago
7

2.c)in the diagram below, 4 m and ORIST at R. 5 $ 0 If m_1 = 63, find m_2.

Mathematics
1 answer:
pickupchik [31]10 months ago
8 0

Data:

l and m are parallel lines

QR and ST are perpendicular in R

Angle 1 is 63°

The angle formed by perpendicular lines is a right angle (90°)

Angles 1 and 3 are alternate angles: angles that occur on opposite sides of a transversal line that is crossing two parallel.

Alternate angles are congruent, have the same measure.

\begin{gathered} m\angle1=m\angle3 \\  \\ m\angle3=63 \end{gathered}

The sum of the interior angles of a triangle is always 180°. In triangle QRT:

\begin{gathered} m\angle2+m\angle3+90=180 \\  \\ m\angle2+63+90=180 \\  \\ m\angle2+153=180 \end{gathered}

Use the equation above to find the measure of angle 2:

\begin{gathered} \text{Subtract 153 in both sides of the equation:} \\ m\angle2+153-153=180-153 \\  \\ m\angle2=27 \end{gathered}Then, the measure of angle 2 is 27°

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2)

P(4,-4) -->(-4, 7)

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C(3,-1) , left 4 up 1

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GuDViN [60]

Answer:

vertex = (0, -4)

equation of the parabola:  y=3x^2-4

Step-by-step explanation:

Given:

  • y-intercept of parabola: -4
  • parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola:  y=a(x-h)^2+k

(where (h, k) is the vertex and a is some constant)

Substitute point (0, -4) into the equation:

\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}

Substitute point (-2, 8) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}

Substitute point (1, -1) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}

Equate to find h:

\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}

Substitute found value of h into one of the equations to find a:

\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}

Substitute found values of h and a to find k:

\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}

Therefore, the equation of the parabola in vertex form is:

\implies y=3(x-0)^2-4=3x^2-4

So the vertex of the parabola is (0, -4)

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