Ryan makes £528 of profit
A) Your primary concerns are the points B and E, so y> .5x+4 and y>or= x-4B) choose one or both points, and enter them into the equations. If the statements are true, then the equations work
for problem C So, any point in the shaded area, but not on the line, are valid points for Natalie's school
11. 6÷5=$1.2 12. 50÷7.90= $6.32 13. 24÷2.95= $8.13 14. 200÷12=$16.6 15. 14.40÷4.5= $3.2 or (4.5÷14.4= 0.3125- Probably not the right answer, I just know it. :)) 16. 22÷2 = $8. Here are your answers! I hope they are right. :)
The first false statement in the proof as it stands is in Line 5, where it is claimed that a line of length 2.83 is congruent to a line of length 4.47. This mistake cannot be corrected by adding lines to the proof.
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The first erroneous tactical move is in Line 4, where the length of DE is computed, rather than the length of FD. This mistake can be corrected by adding lines to the proof.
A correct SAS proof would use segment FD in Line 4, so it could be argued that the first mistake is there.
Answer:
16
Step-by-step explanation:
swtich tk it be 10/160
10 into 160 is 0
subtract 160-0=1
move the 2nd digt down
16-10=6
bring down 3rd digit
=60
60 into 60 is 0