Which steps should be used to graph the equation y- - 1x + 2)?

Mathematics

1 answer:

sweet [91]3 years ago

7 0

Is there any image or anything?

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Is of I ready please help me ?

blsea [12.9K]

Answer: N has to be a negative. hope this helps

Step-by-step explanation:

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3 years ago

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.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago

Simplify and leave in radical form. 8√x^2 y^6

IrinaK [193]

Step-by-step explanation:

$\sqrt[8]{ {x}^{2} {y}^{6} } \\ \\ = ( {x}^{2} {y}^{6} )^{ \frac{1}{8} } \\ \\ = {x}^{2 \times \frac{1}{8}} {y}^{6\times \frac{1}{8}} \\ \\ = x^{\frac{1}{4}} {y}^{3\times \frac{1}{4}} \\ \\ = x^{\frac{1}{4}} {y}^{\frac{3}{4}} \\ \\ = \sqrt[4]{x {y}^{3} } \\$