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Ilia_Sergeevich [38]

4 years ago

12

Find the total surface area for the following. height-3m, length-300 cm and width- 2 cm.

2 answers:

Delvig [45]4 years ago

6 0

Answer:

Step-by-step explanation:

So the Dimensions are:

Height: 300cm

Length: 300cm

Width: 2cm

SA = 2(H*L)+2(H*W)+2(L*W)

= 180000 + 1200 + 1200

= 182400 $cm^{2}$ OR 18.24 $m^{2}$

cluponka [151]4 years ago

6 0

Answer:

182400 cm²

Step-by-step explanation:

At = 2( l×w + l×h + h×w)

= 2(300cm×2cm + 300cm×300cm + 300cm×2cm)

= 2(600cm² + 90000cm² + 600cm²)

= 2×91200cm²

= 182400 cm²

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Find the volume of a cone with a base radius of 6 cm and a height of 10 cm write in terms of pie

Savatey [412]

Answer:

<u>Volume of the cone is 120 π cm³ </u>

Step-by-step explanation:

Given that Base radius of cone is 6 cm and height is 10 cm and we need to find volume of the cone. Volume of cone is given by:

$\\ \: \: \dashrightarrow \: \: \: \underline{\boxed{\pmb{\sf{Volume_{(cone)} = \dfrac{1}{3} \pi r^2 h }}}} \\ \\$

On Substituting the required values, we get:

$\\ \: \: \dashrightarrow \: \: \sf \: Volume_{(cone)} = \frac{1}{3} \times \pi \times {(6)}^{2} \times 10 \\ \\ \\ \: \: \dashrightarrow \: \: \sf \: Volume_{(cone)} = \frac{1} {\cancel{3}} \times \pi \times \cancel{36} \times 10 \\ \\ \\ \: \: \dashrightarrow \: \: \sf \: Volume_{(cone)} = \pi \times 12 \times 10 \\ \\ \\ \: \: \dashrightarrow \: \: \sf \: \: { \purple{Volume_{(cone)} =120 \pi \: cm^3 }}\\ \\$

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6 0

2 years ago

Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position

stira [4]

Answer:

$v(t) = (2t+1) \mathbb{i} +3t^2 \mathbb{j} +4t^3 \mathbb{k}$

$r(t) = (t^2+t) \mathbb{i} +(t^3+2) \mathbb{j} +(t^4 - 3) \mathbb{k}$

Step-by-step explanation:

The velocity vector is the integral of the acceleration vector i.e.

$v(t) = \int a(t) dt$

$v(t) = \int (2 \mathbb{i}+6t \mathbb{j} 12t^2 \mathbb{k}) dt$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + C_1$

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$v(t) = (2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}$

The position vector is the integral of the velocity vector i.e.

$r(t) = \int v(t) dt$

$r(t) = \int ((2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}) dt$

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nata0808 [166]

Answer:

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