Determine the area of the parallelogram with vertices (4, 6, 2), (4, 7, 2), (9, 7, 2), and (9, 8, 2). Use the square root symbol
astra-53 [7]
Answer:
5 square units
Step-by-step explanation:
The parallelogram lies entirely in the plane z=2, so we can treat this as a 2-dimensional problem.
In the order given, the vertices do not define a parallelogram. Two of the sides are parallel, but the other two sides cross each other, for a net area of zero.
If we swap the order of the last two vertices, we get a parallelogram that has a base of 1 unit and a height of 5 units. Its area is ...
A = bh = 1·5 = 5 . . . square units
Answer:
It subtracts
Step-by-step explanation:
Answer:
Step-by-step explanation:
Let the points be A and B
Let A ( - 3 , -2 ) be ( x₁ , y₁ ) and B ( -3 , 6 ) be ( x₂ , y₂ )
<u>Finding</u><u> </u><u>the </u><u>distance</u><u> </u><u>between </u><u>these</u><u> </u><u>points</u>
Hope I helped!
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Answer:
The acceleration with which particle travel after 1 sec is - 2 mps²
Step-by-step explanation:
Given as :
The acceleration of particle moving as a(t) = - 3 t + t²
Now, for time period of acceleration = 1 unit
So, At t = 1
A( 1 ) = - 3 ×( 1 ) + ( 1 )²
Or, A( 1 ) = - 3 + 1
∴ A( 1 ) = - 2 meter per sec²
Hence The acceleration with which particle travel after 1 sec is - 2 mps² Answer