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IceJOKER [234]
3 years ago
8

Help me plz 8th grade math need answers quick

Mathematics
1 answer:
mixer [17]3 years ago
8 0

Answer:

25. Not a solution

26. Yes a solution

27. y= 2/3 x

28. y=3x+16

Step-by-step explanation:

25. The solution to a linear equation is the point (x,y). To find if an (x,y) is a solution, substitute it into the equation and see if it works.

y=4x-4 and (8,3)

3=4(8)-4

3=32-4

3=28

False

Not a solution.

26. Repeat 25 to solve 26 with y=5x-5 and (0,-5)

y=5x-5 and (0,-5)

-5=5(0)-5

-5=0-5

-5=-5

True

Yes a solution.

27. This line crosses through the origin is proportional and therefore has the form y=mx. It also has a gentle slant meaning that is less than 1/2 and goes in a positive direction. This means it is y=2/3 x.

28. Use inverse operations to rearrange the equation.

y-3x=16

y=16+3x

y=3x+16

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