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SVEN [57.7K]
4 years ago
7

What is the solution of 3a + 6 < 21

Mathematics
1 answer:
svlad2 [7]4 years ago
8 0

Answer:

a<5

Step-by-step explanation:

Let's solve your inequality step-by-step.

3a+6<21

Step 1: Subtract 6 from both sides.

3a+6−6<21−6

3a<15

Step 2: Divide both sides by 3.

3a

3

<

15

3

a<5

Answer:

a<5

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Let's factorize it:
\frac{ \frac{1}{x-3}+ \frac{4}{x}  }{ \frac{4}{x} - \frac{1}{x-3} } = \frac{ \frac{x*1}{x(x-3)}+ \frac{4*(x-3)}{(x-3)*x}}{ \frac{(x-3)*4}{(x-3)*x}- \frac{x*1}{x*(x-3)}}= \frac{ \frac{x+4(x-3)}{x(x-3)} }{ \frac{4(x-3)-x}{x(x-3)} }

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\frac{ \frac{x+4(x-3)}{x(x-3)} }{ \frac{4(x-3)-x}{x(x-3)} }= \frac{(x+4(x-3))*x(x-3)}{(4(x-3)-x)*x(x-3)}

Cancel x(x-3) and simplify:
\frac{(x+4(x-3))*x(x-3)}{(4(x-3)-x)*x(x-3)} = \frac{x+4(x-3)}{4(x-3)-x} = \frac{x+4x-12}{4x-12 -x} = \frac{5x-12}{3x-12}
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Or, 2 =  ( 1 + \dfrac{\textrm R}{400})^{\textrm 36}

Or, 2^{\frac{1}{36}} = 1 + \dfrac{\textrm R}{400}

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