Prove that kn=mp and knmp is a parallelogram

Mathematics

1 answer:

____ [38]3 years ago

3 0

Answer:

See the paragraph proof below.

Step-by-step explanation:

Quadrilateral JKLM is given as a parallelogram. By a theorem, opposite sides JK and LM are congruent (1). By the definition of parallelogram, opposite sides KJ and ML are parallel. By the theorem on alternate interior angles, angles KJL and MLJ are congruent (2). Segments JN and PL are given as congruent (3). Using the three statements of congruence labeled above (1), (2), and (3), we now prove that triangles JKN and LMP are congruent by SAS. Sides of the triangles KN and PM are congruent by CPCTC. Sides of quadrilateral KNMP are given as parallel. Therefore, quadrilateral KNMP is a parallelogram by the theorem: If two sides of a quadrilateral are both parallel and congruent, then the quadrilateral is a parallelogram.

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What is the answer what should I write down I will mark you the Brainlyiest 40 points so do it good!

Leni [432]

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3 years ago

Read 2 more answers

A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
<em>Then for the x direction</em> it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, <em>for the y direction</em>, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$