Answer: 56
Step-by-step explanation:
The least common multiples of 7 are 7, 14, 21, 28, 35, 42, 49, and 56.
The least common multipes of 8 are 8, 16, 24, 32, 40, 48, and 56.
56 is the least common multiple of 7 and 8.
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>
So if we want to know the common solution(s) to a system of 2 equations, So we can just set both equations equal to each other and solve for the x value(s). That’s where I start below;
2x^2-13x+21 = 2x^2+9x-56
2x^2 cancels out and moving everything to one side and anything with an x variable to the other side we have then;
-22x=-77
22x=77 by cancelling the negative signs
x=77/22 therefore x=7/2 or 3.5
Hope this helps you. Any questions please ask.