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KatRina [158]
4 years ago
14

If a(x) = 3x + 1 and b(x)= squareroot x-4, what is the domain of (b*a)(x)?

Mathematics
2 answers:
pickupchik [31]4 years ago
5 0

Answer:

c

Step-by-step explanation:

got it on edge

oksano4ka [1.4K]4 years ago
4 0

\boxed{ \ x \geq 1 \ } or can be written as \boxed{ \ [1, \infty) \ }

<h3>Further explanation</h3>

This is a question about the composition of functions and how to get a domain function.

Given \boxed{ \ a(x) = 3x + 1 \ } and \boxed{ \ b(x) = \sqrt{x - 4} \ }.

We will form (b o a)(x) and then determine the domain.

<u>Step-1</u>

\boxed{ \ (b \circ a)(x) = b(a(x)) \ }

Replace each appearance of x in b(x) with \boxed{ \ a(x) = 3x + 1 \ }.

\boxed{ \ (b \circ a)(x) = \sqrt{(3x + 1) - 4} \ }

Thus, \boxed{ \ (b \circ a)(x) = \sqrt{3x - 3} \ }

<u>Step-2</u>

To be defined, the value under the radical sign must not be negative. Therefore, the domain of (b \circ a)(x) = \sqrt{3x - 3} are processed as follows.

\boxed{ \ 3x - 3 \geq 0 \ }

Both sides added by 3.

\boxed{ \ 3x \geq 3 \ }

Both sides divided by 3.

\boxed{ \ x\geq 1 \ }

Thus, the domain of (b \circ a)(x) = \sqrt{3x - 3} is \boxed{ \ x \geq 1 \ } or can be written as \boxed{ \ [1, \infty) \ }

<h3>Learn more</h3>
  1. If f(x) = x² – 2x and g(x) = 6x + 4, for which value of x does (f o g)(x) = 0? brainly.com/question/1774827
  2. Solve for the value of the function composition brainly.com/question/2142762
  3. Look for rotation rules in the transformation brainly.com/question/2992432

Keywords: composition of function, if a(x) = 3x + 1, and, b(x) = √(x-4), what is the domain of, (b o a)(x), b(a(x)), defined, the value, under the radical sign, must not be negative,

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P(t) = [ 64/ (1 + 11.e⁽⁻⁰ ⁰⁸t⁾]
In 1991, t = 1, hence:

P(t) = [ 64/ (1 + 11.e⁽⁻⁰ ⁰⁸ˣ¹⁾]  =  5.7377 billion or rounded 5.74 billion
(Answer A)
5 0
3 years ago
Consider a set of one-dimensional points: 6, 12, 18, 24, 30, 42, 48.
Vinvika [58]

Answer:

1) With initial centroids 10-40, final clusters are

first cluster (6,12,18,24,30)

second cluster (42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 378

2) With initial centroids 10-20, final clusters are

first cluster (6,12,18,24)  

second cluster (30,42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 348

Step-by-step explanation:

K-means works as follows

  • the points will be clustered according to their distance to the centroids.
  • Then centroids are updated as the cluster means.
  • This process continues until clusters doesn't change anymore

1)  <u><em>initial centroids</em></u> 10-40

first cluster (6,12,18,24)  mean:15

second cluster (30,42,48) mean:40

<u><em>new centroids</em></u> 15-40

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

<u><em>final centroids</em></u> 18-45

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

Sum of Squared errors = (18-6)^{2}+(18-12)^{2}+(18-18)^{2}+(18-24)^{2}+(18-30)^{2}+(45-42)^{2}+(45-48)^{2}=378

2)<u><em>initial centroids</em></u> 10-20

first cluster (6,12)  mean:9

second cluster (18,24,30,42,48)  mean:32.4

<u><em>new centroids</em></u> 9-32.4

first cluster (6,12,18) mean:12

second cluster (24,30,42,48) mean:36

<u><em>new centroids</em></u> 12-36

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

<u><em>final centroids</em></u> 15-40

first cluster (6,12,18,24) mean:15  

second cluster (30,42,48) mean:40

Sum of Squared errors = (15-6)^{2}+(15-12)^{2}+(15-18)^{2}+(15-24)^{2}+(40-30)^{2}+(40-42)^{2}+(40-48)^{2}=348

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