Answer:
-3 >v
Step-by-step explanation:
3v-37>-2(8-5v)
Distribute
3v -37> -16 + 10v
Subtract 3 from each side
3v -3v -37> -16 + 10v-3v
-37 > -16 +7v
Add 16 to each side
-37 +16 > -16+16 +7v
-21 > 7v
Divide each side by 7
-21/7 >7v/7
-3 >v
(3x-5)(2x+7) =
= 3x·2x + 3x·7 - 5·2x - 5·7
= 6x² + 21x - 10x - 35
= 6x² + 11x - 35
Ok let’s solve it
5(x-2)^2-20=0
first let’s foil (x-2)
5(x^2-4x+4) -20=0
now distribute the 5
5x^2 -20x +20 -20 = 0
combine like terms
5x^2-20x=0
take the gcf
5x(x-4)=0
x=0, 4
solutions are (4,0) and (2, -20) because the original vertex form a(x-h)^2+k
Answer:
D. 16 ft
Step-by-step explanation:
h = √20² - (24/2)² = √400-144 = √256 = 16
Answer: 63lbs
Step-by-step explanation:
The truck can only hold 16crates and each crate weigh 12lbs, the total weight of the 16 crates will be 12×16= 192lbs
This shows that the truck can only contain extra load of 1200lbs - 192lbs = 1008lbs (excluding weight of crates).
To get the shipment weigh close to the total of 1200lbs, the truck must be loaded with engine components not more than 1008lbs.
Since we have 16 crates to fill with engine components not more than 1008lb, each crates will therefore must not exceed 1008/16 pounds of engine components which is equivalent to 63lbs.
Therefore, the manager should instruct the workers to put 63lbs of machine components in "each crate" in order to get the shipment weight as close as possible to 1200 lbs.
