Answer:
89–100
Step-by-step explanation:
Let x represent the grade on the fifth test. The average for 5 tests wants to be between 80 and 90, so we have ...
80 ≤ (78 +62 +91 +80 +x)/5 ≤ 90
Multiplying by 5 gives
400 ≤ 311 +x ≤ 450
Subtracting 311 gives the range for the fifth test grade.
89 ≤ x ≤ 139
Since the maximum available score is 100, the range on the 5th test for an overall B grade is ...
89 ≤ x ≤ 100 . . . . where x is the grade on the 5th test
Answer:
2k
Step-by-step explanation:
-k+3k
factor out a k
k(-1+3)
k(2)
2k
Step-by-step explanation:
(i) T = { 4, 8, 12}
U = {1,2,3,4,6,8,12,24}
(iii) TUU = {1,2,3,4,6,8,12,24}
(iv) yes, because all the elements in T are also the elements in U.
Answer:
Step-by-step explanation:
XY = 15 MI.
YZ = 8 mi.
Area = A
A = 1/2 (XY)(YZ)
A = 1/2 (15)(8)
A = 1/2 (120)
A = 60
$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$
