The sixth term of an arithmetic sequence is 6
<h3>How to find arithmetic sequence?</h3>
The sum of the first four terms of an arithmetic sequence is 10.
The fifth term is 5.
Therefore,
sum of term = n / 2(2a + (n - 1)d)
where
- a = first term
- d = common difference
- n = number of terms
Therefore,
n = 4
10 = 4 / 2 (2a + 3d)
10 = 2(2a + 3d)
10 = 4a + 6d
4a + 6d = 10
a + 4d = 5
4a + 6d = 10
4a + 16d = 20
10d = 10
d = 1
a + 4(1) = 5
a = 1
Therefore,
6th term = a + 5d
6th term = 1 + 5(1)
6th term = 6
learn more on sequence here: brainly.com/question/24128922
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The square root of a a negative integer is imaginary.
It would still be a negative under a square root if you multiplied it by 2, therefor it will still be imaginary, or I’m assuming as your book calls it, undefined.
2•(sqrt-1) = 2sqrt-1
If you add a number to -1 itself, specifically 1 or greater it will become a positive number or 0 assuming you just add 1. In that case it would be defined.
-1 + 1 = 0
-1 + 2 = 1
If you add a number to the entire thing “sqrt-1” it will not be defined.
(sqrt-1) + 1 = 1+ (sqrt-1)
If you subtract a number it will still have a negative under a square root, meaning it would be undefined.
(sqrt-1) + 1 = 1 + (sqrt-1)
however if you subtract a negative number from -1 itself, you end up getting a positive number or zero. (Subtracting a negative number is adding because the negative signs cancel out).
-1 - -1 = 0
-1 - -2 = 1
If you squared it you would get -1, which is defined
sqrt-1 • sqrt-1 = -1
and if you cubed it, you would get a negative under a square root again, therefor it would be undefined.
sqrt-1 • sqrt-1 • sqrt-1 = -1 • sqrt-1 = -1(sqrt-1)
Sorry if this answer is confusing, I don’t have a scientific keyboard, I’ll get one soon.
Answer:
(x - 6) meters
Step-by-step explanation:
x² - 12x + 36
x² - 2(x)(6) + 6²
(x - 6)² = side²
side = x - 6
Standard form of a linear equation is
Ax + By = C, where A, B, and C are integers.
Add 3 to both sides to get
4x + 4y = 3
