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xeze [42]
4 years ago
13

Pleaseee help!

Mathematics
1 answer:
user100 [1]4 years ago
7 0

Step-by-step explanation:

f(3)=?

f(x)=2x+5

put x=3,

f(3)=2(3)+5=6+5=11

------------

g(2)=?

g(x)=x^2-3

put x=2,

g(2)=2^2-3=4-3=1

-----------------

g(f(-1))=?

g(x)=x^2-3

and f(-1)=2(-1)+5= -2+5=3

so g(f(-1))=3^2-3=9-3=6

----------------

f(g(-1))=?

f(x)=2x+5

g(x)=g(-1)=(-1)^2-3=1-3= -2

f(g(-1))=2(-2)+5= -4+5=1

----------------

g(f(x))=?

g(x)= x^2-3

put x=f(x),

g(f(x))=f(x)^2-3=(2x+5)^2-3=4x+25+20x-3=24x+25

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Multiplying and dividing radical expressions and leaving them in factored form. I am trying to find the best and easy way to get
natka813 [3]

Answer:

\frac{3x + 8}{36 - 2x} / \frac{27x^2 + 72x}{3x^2 - 27} = \frac{(x - 3)(x+3)}{6x(18 - x)}

Step-by-step explanation:

Given

\frac{3x + 8}{36 - 2x} / \frac{27x^2 + 72x}{3x^2 - 27}

Required

Solve

Change / to *

\frac{3x + 8}{36 - 2x} * \frac{3x^2 - 27}{27x^2 + 72x}

Factor out 3

\frac{3x + 8}{36 - 2x} * \frac{3(x^2 - 9)}{3(9x^2 + 24x)}

\frac{3x + 8}{36 - 2x} * \frac{(x^2 - 9)}{(9x^2 + 24x)}

Factorize:

\frac{3x + 8}{36 - 2x} * \frac{(x^2 - 9)}{3x(3x + 8)}

Cancel out 3x + 8

\frac{1}{36 - 2x} * \frac{(x^2 - 9)}{3x}

Factorize:

\frac{1}{2(18 - x)} * \frac{(x^2 - 9)}{3x}

Combine

\frac{x^2 - 9}{2*3x(18 - x)}

\frac{x^2 - 9}{6x(18 - x)}

Express the numerator as a difference of two squares

\frac{(x - 3)(x+3)}{6x(18 - x)}

Hence:

\frac{3x + 8}{36 - 2x} / \frac{27x^2 + 72x}{3x^2 - 27} = \frac{(x - 3)(x+3)}{6x(18 - x)}

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