(30*5)-(10n)>=100
150-(10n)>=100
-(10n)>=100-150
n<=(100-150)/-10
n<=5
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(30 x 5) - 10n ≥ 100
150 - 10n ≥ 100
-10n ≥ 100 - 150
-10n ≥ -50
n ≤ -50 / -10
n ≤ 5
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Here’s how I worked through it
(30 * 5) -10n >= 100
-10n >= 100 -150
n <= -50/-10
n <= 5
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Here’s my working for this one.
(30 x 5) - 10n >= 100
150 - 10n >= 100
-10n >= -50
n <= 50/10
n <= 5
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(30x5)-(10n) >= 100
150-10n >= 100
-10n >= 50
n <= 5
Players can fuel 5 or less number of times and still afford upgrade.
Problem: (30 x 5) - 10n >= 100
30 * 5 = 150
150 - 10n >= 100
150 - 10n >= 100 - 150 (Subtract 150 on both sides)
-10n >= -50
-10*n/-10 >= -50/-10 (Divide with -10 on both sides to cancel out the multiply operation)
Answer: n <= 5
Hello everyone!
Her is how I fare
- Provided the players NEVER MISSES a pickup
( 30 * 5 ) - ( 10 n ) > = 100
-( 10 n ) >= 100 - ( 30 * 5 )
-10n >= 100 - 150
-1 (-10 n) <= -1( 100 - 150 ) … WE FLIP THE SIGN HERE
10 n <= -100 + 150
n <= 50/10
[ n <= 5 ] - So our players have to refuel 5 times OR LESS to afford an upgrade
30x5-(10n)>=100
150-(10n)>=100
-10n>=-50
n<=5
(30x5) - (10n) >= 100 becomes 10(15 - n) >= 100 becomes (15-n) >= 10 becomes n<= 5
(30 * 5) – (x * 10) ≥ 100
150 – (10x) ≥ 100
-(10x) ≥ -50
10x ≤ 50
x ≤ 5
(30 * 5) - (10 * n) >= 100
150 - 10n >= 100
150 >= 100 + 10n
50 >= 10n
5 >= n
n <= 5
(30 * 5) - (10n) >= 100
-10x >= 100 - (30 * 5)
-10x >= 100 - 150
-10x >= -50
x <= -50 / -10
x <= 5
(30*5) - (10n) >= 100
150 - (10n) >= 100
-10n >= -50
n <= 5
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30 * 5 - 10X >= 100
150 - 10X >= 100
-10 * X >= -50 (divide both by -10)
X <= 5```
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(30*5) - (10n) >= 100 #Original problem.
150 - 10n >= 100 #Removed parenthesis.
-10n >= -50 #Subtract 150 from both sides.
n <= -50/-10 #Divide both sides by -10 and flipped sign of inequality.
n <= 5 #Simplified answer.
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Hello. I’m not understanding the Upgrade and why one would use it?
Do you need the Upgrade to Refuel your ship?
Does the Upgrade increase the Fuel Tank Size (max fuel size increase)?
Left blank for spoilers.
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I only see this issue as:
a=(30x5) (a/10) which is 15 times the player can refuel
The Upgrade will not be available when a <= 100 (at steps of 10 this would be 5 Refuels)
My questions may seem strange, I’m not into Math(s)
Thanks
Ps. how do you use spoiler tags?
Here’s my answer for the inequalities challenge:
Q1. Given (30 * 5) - 10n >= 100, solve for n
A1. The player can refuel a total of five times before they can no longer afford an upgrade.
A1a. (30 * 50) - 10n >= 100
A1b. 150 - 10n >= 100
A1c. -10n >= -50
A1d. n <= 5
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So here it is…
(30 * 5) - (10n) >= 100
150 - 10n >= 100
-10n >= 100 - 150
-10 >= -50 → both sides * (-10)
n <= 5
Player can refuel maximum 5 times, and still can buy an upgrade.
Check
(30 * 5) - (10 * 5) >= 100
150 - 50 >= 100
100 >= 100
Looks good
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Hi,
I created this very simple program, the user inputs their budget (upgrade) and how many times they need the pickup, and the program calculates the maximum number of refuel times. Just a small question out of this topic, why I can not ceil (always round up) the numberofrefuel variable?
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When does it not work? It should be possible 