<h3>
Answer: Largest value is a = 9</h3>
===================================================
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.
------------------
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
Answer:
<h3>Daniel is correct</h3>
Step-by-step explanation:
Given the polynomial P(x) = x^2-3x-10
To check that the remainder is zero if divided by x - 5, we will first have to equate x - 5 to zero and get x;
x - 5 = 0
x = 5
Then find P(5)
P(5) = 5^2 - 3(5) - 10
P(5) = 25 - 15 - 10
P(5) = 25-25
P(5) = 0
<em>This shows that x - 5 is a factor of the polynomial since P(5) gave us zero according to the factor theorem</em>
Answer:
No.
Step-by-step explanation:
The numbers aren't going up at a constant rate. Say each lap costs 2 dollars. On lap 4, you have paid 8 dollars total. On lap 9, you have paid 18 dollars total. But the 10th lap is free, which means you pay 0 dollars. In order for this pattern to be constant, you would need to pay 20 dollars, but instead you pay 0.
Answer:
C) The sum of the first 2 terms after the fourth term.
Answer:
rational
Step-by-step explanation:
