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3 years ago

ASAP 6(a + 25 + 3c) =

Mathematics

1 answer:

Blizzard [7]3 years ago

7 0

Answer:

6a+150+18c

Step-by-step explanation:

6(a+25+3c)

6a +6 x 25 +6 x 3c

6a + 150+6 x 3c

6a +150 +18c

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3x-6<12 step by step equation

Mandarinka [93]

Answer:

x < 6

Step-by-step explanation:

Given

3x - 6 < 12 ( add 6 to both sides )

3x < 18 ( divide both sides by 3 )

x < 6

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3 years ago

Subtract the rational expression 1/x^2-x-6 - x/x+2.

Aleksandr [31]

Answer:

Step-by-step explanation:

x² - x -6 = x² + 2x - 3x - 3*2

=x(x + 2) - 3(x + 2)

= (x + 2 )(x - 3)

$\frac{1}{x^{2}-x-6}+\frac{x}{x+2}=\frac{1}{(x +2 )(x - 3)}-\frac{x}{x + 2}\\$

$= \frac{1}{(x+2)(x-3)}-\frac{x(x-3)}{(x+2)(x-3)}\\\\=\frac{1-x(x-3)}{(x+2)(x-3)}\\\\=\frac{1-x^{2}+3x}{(x+2)(x-3)}\\\\$

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2 years ago

What is 1/10 of 30000

nevsk [136]

1/10 of 30000 is 3000 because 30000 divided by 10 is 3000.

8 0

3 years ago

Read 2 more answers

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.