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kupik [55]
3 years ago
9

Martha drew a pair of intersecting non-perpendicular lines, I and m. She numbered one pair of vertical angles <1 and <2, a

nd she labeled the second pair of vertical angles <3 and <4. Martha made the conjecture that <1 =~ <2 to validate this conjecture, which of the following would be appropriate reasoning?
A. M<1+ M<2 +M<3 + m<4 = 360

B. The sum of the measures of angles of a triangle is 180°

C. The measure of all right is 90°

D. Because <1 and <2 are each supplementary to <3, they are therefore congruent
Mathematics
1 answer:
RUDIKE [14]3 years ago
8 0

Answer:

  • D. Because <1 and <2 are each supplementary to <3, they are therefore congruent

Step-by-step explanation:

A. m<1 + m<2 + m<3 + m<4 = 360

  • Incorrect in terms of the proof

B. The sum of the measures of angles of a triangle is 180°

  • Not relevant

C. The measure of all right is 90°

  • Not relevant

D. Because <1 and <2 are each supplementary to <3, they are therefore congruent

  • Correct
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fomenos

Answer:

The time taken for the upward motion is 1 second. The same time is taken for the downward motion

It reaches a maximum height of 4.9 meters.

Step-by-step explanation:

The equation of motion is:

x(t) = -4.9t^{2} + 9.8t

Since the term which multiplies t squared is negative, the graph is concave down, that is, x increases until the vertex, where it reaches it's maximum height, then it decreases.

Vertex of a quadratic equation:

Quadratic equation in the format x(t) = at^{2} + bt + c

The vertex is the point (t_{v}, x(t_{v})), in which

t_{v} = -\frac{b}{2a}

In this question:

x(t) = -4.9t^{2} + 9.8t

So a = -4.9, b = 9.8

Vertex:

t_{v} = -\frac{9.8}{2*(-4.9)} = 1

The time taken for the upward motion is 1 second.

x(t_{v}) = x(1) = 9.8*1 - 4.9*(1)^{2} = 4.9

It reaches a maximum height of 4.9 meters.

Downward motion:

From the vertex to the ground.

The ground is t when x = 0. So

-4.9t^{2} + 9.8t = 0

4.9t^{2} - 9.8t = 0

4.9t(t - 2) = 0

4.9t = 0

t = 0

Or

t - 2 = 0

t = 2

It reaches the ground when t = 2 seconds.

The downward motion started at the vertex, when t = 1.

So the duration of the downward motion is 2 - 1 = 1 second.

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