Hi! to solve the question, how many necklaces did anne make in 16 hours if she makes 10 items in 2 standard 8-hour work days, we should first check the time she needs to make each item.
3 individual hours to make a necklace, and 1 hour to make a ring. according to the problem, she made 10 items in 16 hours. this problem will most likely be trial and error. let’s try one first guess.
my first guess will be, maybe she will make 6 rings and 4 necklaces. 6 times 1 is 6, which takes 6 hours for the rings. 4 times 3 is equal to 12. 6 + 12 is too high! let’s try again.
what about 6 rings and 2 necklaces? 6 times 1 is 6, so that’s 6 hours. 2 times 3 is 6 hours! 6 plus 6 is 12, which is close, but not quite there!
let’s try 7 rings and 3 necklaces! 7 times 1 is 7 hours. 3 times 3 is equal to 9 hours! total that together, 7 plus 9, is equal to 16 hours spent!
so, the answer would be, “Anne made 7 rings and 3 necklaces within 16 hours.” hope this helped!:)
Answer:
p(2p-11q+24p^6)
Step-by-step explanation:
Answer:
y = -4
Step-by-step explanation:
3y - 2(5y - 7) = -9y + 6 (Given)
3y - 10y + 14 = -9y + 6 (Distribute)
-7y + 14 = -9y + 6 (Add like terms)
2y + 14 = 6 (Add 9y on both sides)
2y = -8 (Subtract 14 on both sides)
y = -4 (Divide 2 on both sides)
You can check by substituting the solution for y:
3(-4) - 2(5(-4) - 7) = -9(-4) + 6
-12 - 2(-20 - 7) = 36 + 6
-12 - 2(-27) = 42
-12 + 54 = 42
42 = 42
Answer:
24 miles UwU
Step-by-step explanation: G'day mate!
The solution for the given equations x + 3y = 5 and x - 3y = -1 are x=2 and y=1.
Step-by-step explanation:
The given is,
......................................(1)
....................................(2)
Step:1
By elimination method,
Subtracting Equations (1) and (2),
( - )
( 
) + ( 
) = ( 
)
= 6
= 1
Step:2
Substitute the value of y in equation (1),
+ ( 3 × 1 ) = 5
= 5-3
= 2
Step:3
Check for Solution,
Substitute the values of x and y in Eqn (1),
2 + ( 3 × 1 ) = 5
5 = 5
Result:
The solution for the given equations x + 3y = 5 and x - 3y = -1 are x=2 and y=1, by elimination method.
