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 answer is A! This is because the formula is (x-h)^2 + (y-k)^2 = r^2.
Because there is no value, you can see that h and k are zero. This means that the center, (h,k) is (0,0).
So you will need to solve for x and y before evaluating 2x+y....
2x-y=9, y=2x-9 now this will make 4x^2-y^2=171 become:
4x^2-(2x-9)^2=171
4x^2-(4x^2-36x+81)=171
36x-81=171
36x=252
x=7, now we can use 2x-y=9 to solve for y...
2(7)-y=9
14-y=9
-y=-5
y=5
now we know that x=7 and y=5, 2x+y becomes:
2(7)+5
14+5
19
Answer:
2100
Step-by-step explanation: