Pls no links, I really need help
2 answers:
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Greetings!
This can be solved by creating a simple equation.
Solve for x.
Find the LCM.
Add -5/12 to both sides.
Simplify.
Find the LCM.
He kept 7/12 of the raisins for himself.
Hope this helps.
-Benjamin
This can be solved by creating a simple equation.
Solve for x.
Find the LCM.
Add -5/12 to both sides.
Simplify.
Find the LCM.
He kept 7/12 of the raisins for himself.
Hope this helps.
-Benjamin
<h3>Answer: Largest value is a = 9</h3>
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Work Shown:
b = 5
(2b)^2 = (2*5)^2 = 100
So we want the expression a^2+3b to be less than (2b)^2 = 100
We need to solve a^2 + 3b < 100 which turns into
a^2 + 3b < 100
a^2 + 3(5) < 100
a^2 + 15 < 100
after substituting in b = 5.
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Let's isolate 'a'
a^2 + 15 < 100
a^2 < 100-15
a^2 < 85
a < sqrt(85)
a < 9.2195
'a' is an integer, so we round down to the nearest whole number to get
So the greatest integer possible for 'a' is a = 9.
------------------
Check:
plug in a = 9 and b = 5
a^2 + 3b < 100
9^2 + 3(5) < 100
81 + 15 < 100
96 < 100 .... true statement
now try a = 10 and b = 5
a^2 + 3b < 100
10^2 + 3(5) < 100
100 + 15 < 100 ... you can probably already see the issue
115 < 100 ... this is false, so a = 10 doesn't work