Eight more than the product of a number and 9 is 6 translate this into an equation
1 answer:
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<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
Answer:
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Algebra I</u>
- Average Rate of Change:
Step-by-step explanation:
<u>Step 1: Define</u>
Interval -1 ≤ x ≤ 3
a = -1, b = 3
f(a) = f(-1) = 4
f(b) = f(3) = -4
<u>Step 2: Find Average</u>
- Substitute in variables [ARC]:
- Substitute:
- [Fraction] Subtract/Add:
- [Fraction] Divide: