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KatRina [158]
3 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]3 years ago
5 0

Answer:

c

Step-by-step explanation:

got it on edge

oksano4ka [1.4K]3 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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Lines 3x-2y+7=0 and 6x+ay-18=0 is perpendicular. What is the value of a?
BlackZzzverrR [31]

Answer:

\boxed{\sf a = 9 }

Step-by-step explanation:

Two lines are given to us which are perpendicular to each other and we need to find out the value of a . The given equations are ,

\sf\longrightarrow 3x - 2y +7=0

\sf\longrightarrow 6x +ay -18 = 0

Step 1 : <u>Conver</u><u>t</u><u> </u><u>the </u><u>equations</u><u> in</u><u> </u><u>slope</u><u> intercept</u><u> form</u><u> </u><u>of</u><u> the</u><u> line</u><u> </u><u>.</u>

\sf\longrightarrow y = \dfrac{3x}{2} +\dfrac{ 7 }{2}

and ,

\sf\longrightarrow y = -\dfrac{6x }{a}+\dfrac{18}{a}

Step 2: <u>Find </u><u>the</u><u> </u><u>slope</u><u> of</u><u> the</u><u> </u><u>lines </u><u>:</u><u>-</u>

Now we know that the product of slope of two perpendicular lines is -1. Therefore , from Slope Intercept Form of the line we can say that the slope of first line is ,

\sf\longrightarrow Slope_1 = \dfrac{3}{2}

And the slope of the second line is ,

\sf\longrightarrow Slope_2 =\dfrac{-6}{a}

Step 3: <u>Multiply</u><u> </u><u>the </u><u>slopes </u><u>:</u><u>-</u><u> </u>

\sf\longrightarrow \dfrac{3}{2}\times \dfrac{-6}{a}= -1

Multiply ,

\sf\longrightarrow \dfrac{-9}{a}= -1

Multiply both sides by a ,

\sf\longrightarrow (-1)a = -9

Divide both sides by -1 ,

\sf\longrightarrow \boxed{\blue{\sf a = 9 }}

<u>Hence </u><u>the</u><u> </u><u>value</u><u> of</u><u> a</u><u> </u><u>is </u><u>9</u><u> </u><u>.</u>

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