First see how many times can 8 go into 52. So 8 goes 6 times, and it equals to; 48
Now subtract 48 from 52; 52-48= 4
Finally bring down the 0 next to the 4; 40
Now to get you final answer find out how many times can 8 go into 40; 5
Since it goes exactly 5 times then your final answer would be; 65
Hope this helps! :)
Answer:
<h2>10 volunteers are more than recommended</h2>
Step-by-step explanation:
The recommended number of volunteers is five (5)
Since the the distance of the race is 5km,
and the safety committees recommends 1 volunteer per kilometre.
Hence ten (10) volunteers is more than enough
Answer:
11
Step-by-step explanation:
11,16,12,7,23,9,5,5
arrange to least to greatest
5,5,7,9,11 ,12,16,23
2 middle numbers
11+12= 23
23÷2= 11.5
Median is 11
I tried but see if this helps you In algebra, it is easy to find the third value when two values are given. Generally, the algebraic expression should be any one of the forms such as addition, subtraction, multiplication and division. To find the value of x, bring the variable to the left side and bring all the remaining values to the right side.
Answer:
20. AB = 42
21. BC = 28
22. AC = 70
23. BC = 20.4
24. FH = 48
25. DE = 10, EF = 10, DF = 20
Step-by-step explanation:
✍️Given:
AB = 2x + 7
BC = 28
AC = 4x,
20. Assuming B is between A and C, thus:
AB + BC = AC (Segment Addition Postulate)
2x + 7 + 28 = 4x (substitution)
Collect like terms
2x + 35 = 4x
35 = 4x - 2x
35 = 2x
Divide both side by 2
17.5 = x
AB = 2x + 7
Plug in the value of x
AB = 2(17.5) + 7 = 42
21. BC = 28 (given)
22. AC = 4x
Plug in the value of x
AC = 4(17.5) = 70
✍️Given:
AC = 35 and AB = 14.6.
Assuming B is between A and C, thus:
23. AB + BC = AC (Segment Addition Postulate)
14.6 + BC = 35 (Substitution)
Subtract 14.6 from each side
BC = 35 - 14.6
BC = 20.4
24. FH = 7x + 6
FG = 4x
GH = 24
FG + GH = FH (Segment Addition Postulate)
(substitution)
Collect like terms


Divide both sides by -3

FH = 7x + 6
Plug in the value of x
FH = 7(6) + 6 = 48
25. DE = 5x, EF = 3x + 4
Given that E bisects DF, therefore,
DE = EF
5x = 3x + 4 (substitution)
Subtract 3x from each side
5x - 3x = 4
2x = 4
Divide both sides by 2
x = 2
DE = 5x
Plug in the value of x
DE = 5(2) = 10
EF = 3x + 4
Plug in the value of x
EF = 3(2) + 4 = 10
DF = DE + EF
DE = 10 + 10 (substitution)
DE = 20