Ihave a couple questions with my geometry homework which is due tomorrow, could someone try answering them? i've been trying to solve them for the past hour.

  2.  100°              22.  7

  18.  6                  23.  9

  19.  8                  24.  65°

  20.  55°             25.  AB = (1/2)DF

  21.  6                 26.  AB ║ DF

Step-by-step explanation:

Please be aware that the triangle measurements shown for problems 18–20 and 22–24 cannot exist. In the first case, the angle is closer to 48.6° (not 35°), and in the second case, the angle is closer to 51.1°, not 25°. So, you have to take the numbers at face value and not think too deeply about them. (This state of affairs is all too common in geometry problems these days.)

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2. You have done yourself no favors by marking the drawing the way you have. Look again at the given conditions. You will find that x+2x must total a right angle, so x=30°. Angle P is the complement of 40°, so is 50°. Then the sum of x and angle P is 30° +50° = 80°, and the angle of interest is the supplement of that, 100°.

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18–20. Perpendicular bisector NO means  ∆NOL ≅ ΔNOM. Corresponding parts have the same measures, and angle L is the complement of the marked angle.

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21. ∆ODA ≅ ∆ODB by hypotenuse-angle congruence (HA), so corresponding parts are the same measure. DB = DA = 6.

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22–24. ∆VPT ≅ ∆VPR by LL congruence, so corresponding measures are the same. Once again, the angle in question is the complement of the given angle.

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25–26. You observe that A is the midpoint of DE, and B is the midpoint of FE, so AB is what is called a "midsegment." The features of a midsegment are that it is ...

52 \43 i don't know this i tried