The class 4S because you divide each number by its mean mark and standard division and 4S is the highest.
Answer:
-4.01, -4, -7
Step-by-step explanation:
equation:
-x + 6 ≥ 10
substitute each number for x
- (-3.9) + 6 ≥ 10 = (negatives cancel each other out) 3.9 + 6 = 9.9 ≥ 10 (incorrect)
-4 + 6 ≥ 10 = 2 ≥ 10 (incorrect)
- (-4.01) + 6 ≥ 10 = (negatives cancel each other out) 10.01 ≥ 10 (correct)
- (-4) + 6 ≥ 10 = (negatives cancel each other out) 10 ≥ 10 (correct)
-4.01 + 6 ≥ = 1.99 ≥ 10 (incorrect)
-3.9 + 6 ≥ 10 = 2.1 ≥ 10 (incorrect)
-0 + 6 ≥ 10 = 6 ≥ 10 (incorrect)
- (-7) + 6 ≥ 10 = (negatives cancel each other out) 11 ≥ 10 (correct)
#30 and #31 are in y = mx + b. m = slope, and b = y-intercept.
30. slope = 2/3 y-intercept = 1
31. Slope = -9/5 y-intercept = -4
To make the x's get on the right side of the equal sign, subtract it on both sides.
32. 6x - y = -2
32. 6x - y - 6x = -2 - 6x
32. y = 6x - 2
Answer:
segment EG over segment LN equals segment FG over segment MN
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent
In this problem
The corresponding sides are
EF and LM
EG and LN
FG and MN
The corresponding angles are
∠E≅∠L
∠F≅∠M
∠G≅∠N
therefore
EF/LM=EG/LN=FG/MN=3/1
Answer:
- premium: 165 gallons
- water: 120 gallons
Step-by-step explanation:
Let 'a' represent the amount of 95% antifreeze in the mix. Then the amount of water is (285 -a). The mix has this amount of antifreeze in it:
0.95a +0(285 -a) = 0.55(285)
a = 0.55/0.95(285) = 165 . . . gallons of 95%
285 -a = 285 -165 = 120 . . . . gallons of water
There are 120 gallons of water and 165 gallons of premium antifreeze in the mixture.
