A. The area of the paralelogram is base • height so is (4x-2)•(2x-1)= 8x^2. -8x+2
b. X=2
A= 8•4-8•2+2= 32-16+2= 14 in^2
Let's think about this. MQ is given to be a length of 24 units, PR a length of 10 whilst we must determine what length PM must be in order to satisfy the criteria of parallelogram MPQR to be a rhombus.
Assume this figure is a rhombus, rhombus MPQR. If that is so, all sides must be congruent, and the diagonals must be perpendicular ( ⊥ ) by " Properties of a Rhombus. " That would make triangle( s ) MRQ and say RMP isosceles, and by the Coincidence Theorem, MS ≅ QS, and RS ≅ PS. Therefore -
PS and MS are legs of a right triangle, so by Pythagorean Theorem we can determine the hypotenuse, or in other words the length of PM. This length would make parallelogram MPQR a rhombus,
<u><em>And thus, PM should be 13 in length to make parallelogram MPQR a rhombus.</em></u>
The equation of the line on the xy-coordinate graph ... is the distance from the center of the base to a point on the circumference of the top of the can?
Answer:
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