1. (02.01) Solve -4(x + 10) - 6 = -3(x - 2). (1 point) -40 -46 -52 52

Mathematics

1 answer:

aalyn [17]3 years ago

6 0

Answer:

-52

Step-by-step explanation:

-4(x + 10) - 6 = -3(x - 2)

Distribute the left side to get:

(-4x + -40) - 6

Now distribute the right side to get:

-3x + 6

Arrange the equation as the following:

-4x - 40 - 6 = -3x + 6

Add the like terms on each side:

-4x - 46 = -3x + 6

Do the inverse operation of each term:

-x = 52

Now we need to get x to become a positive, so we just divide -x by -1 to get x.

And 52/-1 to get our final answer of -52.

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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.