Okay first of all #2 is Input is 0 divided by 8=0,16 divided by 8=2,24 divided by 8=3,and 32 divided by 8=4.
Next one,34-4=30,34-8=26,34-12=22,34-16=18,34-20=14,34-24=10,and 34-28=6
4 row 60-47=13,65-47=18,70-47=23,75-47=29,80-47=33,85-47=39
5 row 7+4=11 14+4=18 21+4=25 28+4=32 35+4=39 42+4=46 49+4=53
6 row 1x3=3 2x3=6 3x3=9 4x3=12 5x3=15
7 row 18÷6=3 24÷6=4 30÷6=5 36÷6=6 42÷6=7 48÷6=8 54÷6=9
hope this helps ^-^
If I'm wrong,or make you get the answer wrong Vannellope2 I am so sorry
Answer:
6000 in³
Step-by-step explanation:
To solve this problem, we simply have to find the volume of the shipping container that will be just enough to contain the 20 soda boxes.
To do this, we find the volume of each soda box and multiply it by the total number of soda boxes held by the shipping container.
Volume of the box = L * B * H
L = length = 15 in
B = breadth = 4 in
H = height = 5 in
V = 15 * 4 * 5 = 300 in³
This is the volume of each soda box.
The volume of 20 soda boxes will then be:
V = 20 * 300 = 6000 in³
This is the volume of 20 soda boxes and hence, the minimum size the shipping container can be.
-1/2+2 (-9/4)
Reduce the fraction by dividing the numerator and denominator by 2
-1/2+1 (-9/2)
Multiply the values
-1/2 - 9/2
Bring the values to the common denominator
(-1-9)/2
Calculate the difference
-10/2
Reduce the fraction by dividing the numerator and denominator by 2
-5.
Answer:
C The system has no solution
Step-by-step explanation:
They are parallel lines so there is no solution :).
