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1. 3h + 5t ≤ 500; if h=10
2. d + 2m ≥ 24 if m=6
3. r + b ≤ 100 if r=47
Answers
Step-by-step explanation:
(1. ) 3h + 5t = 500
h = 10
3(10) + 5t = 500
5t = 500 - 30
t = 470/5 = 94
2) d + 2m = 24
m = 6
d + 2(6) = 24
d = 24 - 12 = 12
3 ) r + b = 100
r = 47
47 + b = 100
b = 100 - 47 = 53
Step-by-step explanation:
Given :-
1. 3h + 5t ≤ 500
2. d + 2m ≥ 24
3. r + b ≤ 100
To find:-
Solve the following inequations
1. 3h + 5t ≤ 500; if h=10
2. d + 2m ≥ 24 if m=6
3. r + b ≤ 100 if r=47
Solution:-
1)Given in-equation is 3h+5t≤500
and h=10
substituting the value of h in the given in-equation then
=>3(10)+5t≤500
=>30+5t≤500
on subtracting 30 both sides
=>30-30+5t≤500-30
=>0+5t≤470
=>5t<470
On dividing by 5 both sides
=>5t/5≤470/5
=>t≤94
The maximum value of t=94
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2)Given in-equation is d+2m≥24
and m=6
substituting the value of m in the given in-equation then
=>d+2(6)≥24
=>d+12≥24
on subtracting 12 both sides
=>d+12-12≥24-12
=>d+0≥12
=>d≥12
The minimum value of d=12
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3)Given in-equation is r+b≤100
and r=47
substituting the value of r in the given in-equation then
=> 47+b≤100
On Substituting 47 both sides
=>47-47+b≤100-47
=>0+b≤53
=>b≤53
The maximum value of b=53
_____________________________
Answer:-
1)The maximum value of t=94
2)The minimum value of d=12
3)The maximum value of b=53