Part a.
The domain is the set of x values such that 
, basically x can be equal to -1/2 or it can be larger than -1/2. To get this answer, you solve 
for x (subtract 1 from both sides; then divide both sides by 2). I set 2x+1 larger or equal to 0 because we want to avoid the stuff under the square root to be negative.
If you want the domain in interval notation, then it would be 
which means the interval starts at -1/2 (including -1/2) and then it stops at infinity. So technically it never stops and goes on forever to the right.
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Part b.
I'm going to use "sqrt" as shorthand for "square root"
f(x) = sqrt(2x+1)
f(10) = sqrt(2*10+1) ... every x replaced by 10
f(10) = sqrt(20+1)
f(10) = sqrt(21)
f(10) = 4.58257569 which is approximate
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Part c.
f(x) = sqrt(2x+1)
f(x) = sqrt(2(x)+1)
f(x+2a) = sqrt(2(x+2a)+1) ... every x replaced by (x+2a)
f(x+2a) = sqrt(2x+4a+1) .... distribute
we can't simplify any further
Answer:
Third option
Step-by-step explanation:
We can't factor this so we need to use the quadratic formula which states that when ax² + bx + c = 0, x = (-b ± √(b² - 4ac)) / 2a. However, we notice that b (which is 6) is even, so we can use the special quadratic formula which states that when ax² + bx + c = 0 and b is even, x = (-b' ± √(b'² - ac)) / a where b' = b / 2. In this case, a = 1, b' = 3 and c = 7 so:
x = (-3 ± √(3² - 1 * 7)) / 1 = -3 ± √2
The function that best defines this arithmetic sequence is:
<h3>What is an arithmetic sequence?</h3>
- In an <em>arithmetic sequence,</em> the <u>difference between consecutive terms is always the same</u>, called common difference d.
The recursive function that defines the sequence is:
- In which
is the first term.
In this problem, the sequence is: {9, 5, 1, -3}
- Each term is the previous term subtracted by 4, hence
.
- The first term is 9, hence
.
Hence, the function is:
You can learn more about arithmetic sequences at brainly.com/question/6561461
Answer: 14,000 i think
Step-by-step explanation:
