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natita [175]
2 years ago
13

Calculate the total surface area of a cuboid with the following dimensions. 4m by 6m by 8m​

Mathematics
1 answer:
bonufazy [111]2 years ago
6 0

Answer:

V = 192 m^3

Step-by-step explanation:

The volume of a cuboid is given by

V = l*w*h

V = 4m * 6m *8m

V = 192 m^3

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kobusy [5.1K]

Answer:

C.)-5

Step-by-step explanation:

y2-y1/x2-x1

(-9,6) (-6,-9)

-9-6=-15

-6-(-9)=3

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4 0
3 years ago
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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
What is 3.1666666667 as a fraction
never [62]
<span> 3 </span>1666666667⁄<span>10000000000

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7 0
3 years ago
Which equations represent the line that is parallel to 3x − 4y = 7 and passes through the point (−4, −2)? Check all that apply.
olga55 [171]
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4 0
3 years ago
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Solnce55 [7]

Answer:

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Step-by-step explanation:

I attached a picture of my work and I stuck it into my calculator to check it

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