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2 years ago

3.

Mathematics

1 answer:

iren2701 [21]2 years ago

4 0

The answer is: y = |x + 6|

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Which of the following options have the same value as 60% percent of 94?

elena55 [62]

Answer:

54.4

Step-by-step explanation:

8 0

2 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

A square park is 75 m long. Find the area of the park.<br>

Inga [223]

5625 m^2

If is is square, all sides must be congruent (all sides are 75m). To find area, you multiply width x length (how long). Therefore:

75m x 75m = 5626 m^2

3 0

2 years ago

What is the value of t?<br> 40°<br> 39°

kramer

Answer:41

Step-by-step explanation:

because the 49 degrees and t equal a right angle which is 90 degrees you subtract 49 from 90 which makes the remaining degree aka t 41

5 0

2 years ago

Can someone help please

Anon25 [30]

Answer:

x1.20

y0.75)

Step-by-step explanation:

you have to simplify

3 0

2 years ago