<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.
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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
123 5/10 becuase everything on the left side of the dcimal is a whole number and everything on the right side is a fraction. The 5 is in the 10 place so the denominator would be 10 and the numerator would be 5. Your answer is 123 5/10 or simplified, 123 1/2
Answer:
11/12
Step-by-step explanation:
I took 1.9 and turned it into a fraction and the fraction is 1 9/12 so if you have 11/12 and 1 9/12 then you bought 10/12 more pounds of banana than grapes
The distance is 2.23606 (2.24 rounded)
Answer:
1
Step-by-step explanation:
3+2-(6-(1+1)
3+2-(6-2)
3+2-4
1