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Anit [1.1K]
3 years ago
10

The midpoints of an irregular quadrilateral ABCD are connected to form another quadrilateral inside ABCD. Complete the explanati

on of why the quadrilateral is a parallelogram.​
Mathematics
1 answer:
Juliette [100K]3 years ago
5 0

Answer:

Suppose: M, N, P, Q are the midpoints of AB, BC, CD, AD respectively

=> MNPQ is the quadrilateral inside ABCD

connect B to D, ΔABD has : M is the midpoint of AB

                                               Q is the midpoint of AD

=> MQ is the midpoint polygon of ΔABD

=> MQ // BD and MQ = 1/2.BD (1)

ΔBCD has: N is the midpoint of BC

                   P is the midpoint of DC

=> NP is the midpoint polygon of ΔBCD

=> NP // BD and NP = 1/2.BD    (2)

from (1) and (2) => MQ // NP ( //BD)

                             MQ = NP  (=1/2.BD)

=> MNPQ is a parallelogram.​

=>  the quadrilateral inside ABCD is a parallelogram.​

Step-by-step explanation:

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During halftime of a basketball ​game, a sling shot launches​ T-shirts at the crowd. A​ T-shirt is launched from a height of 4 f
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(a) The time the T-shirt takes to maximum height is 2 seconds

(b) The maximum height is 68 ft

(c) The range of the function that models the height of the T-shirt over time given above is 4 + 64\cdot t - 16 \cdot  t^{2}

Step-by-step explanation:

Here, we note that the general equation representing the height of the T-shirt as a function of time is

h = h_1 + u\cdot t - \frac{1}{2} \cdot g  \cdot  t^{2}

Where:

h = Height reached by T-shirt

t = Time of flight

u = Initial velocity = 64 ft/s

g = Acceleration due to gravity (negative because upward against gravity) = 32 ft/s²

h₁ = Initial height of T-shirt = 4 ft

(a) The maximum height can be found from the time to maximum height given as

v = u - gt

Where:

u = Initial velocity = 64 ft/s

v = Final upward velocity at maximum height = 0 m/s

g = 32 ft/s²

Therefore,

0 = 64 - 32·t

32·t = 64 and

t = 64/32 = 2 seconds

(b) Therefore, maximum height is then

h = 4 + 64\times 2 - \frac{1}{2} \times 32  \times  2^{2}

∴ h = 68 ft

The T-shirt is then caught 41 ft above the court on its way down

(c) The range of the function that models the height of the T-shirt over time given above is derived as

h = h_1 + u\cdot t - \frac{1}{2} \cdot g  \cdot  t^{2}

With u = 64 ft/s

g = 32 ft/s² and

h₁ = 4 ft

The equation becomes

h =4 + 64\cdot t - \frac{1}{2} \times 32  \cdot  t^{2} = 4 + 64\cdot t - 16 \cdot  t^{2}.

6 0
3 years ago
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