Ask question

Ask question

All categories

3 years ago

12

Step by step please and thank you

2 answers:

cupoosta [38]3 years ago

8 0

Answer:

2926 m³

Step-by-step explanation:

<u>Given</u>

Length- 19 metres

Breadth- 14 metres

Height- 11 metres

.

As we know,

Volume of a cuboid = l×b×h

So, Volume = 19×14×11

= 2926 m³

alexira [117]3 years ago

6 0

Well, we have the three dimensions of the box given, the length that is 19 meters, the breadth that is 14 meters and the height that is 11 meters.
Now, the volume = length X breadth X height
Therefore, the volume of the box will be 19 X 14 X 11 = 2926 meter cube.
Really hope that was helpful.

You might be interested in

The area of a library table top is 27 1/8 Square feet the table is 3 1/2 feet wide what is the length of the table

zhuklara [117]

Answer:

7.75 or 7 3/4

Step-by-step explanation:

Divide 27 1/8 by 3 1/2 to find the length, because Length x Width = Area

When you divide, you get 7.75

This is the length.

You can also convert it into fraction form and get 7 3/4

6 0

4 years ago

A salmon was 5 1/2 feet below sea level. It swarm 1 1/4 ft up to get away from a bear. Where is the salmon now located? Show you

castortr0y [4]

Answer:

4 1/4 feet below sea level

Step-by-step explanation:

5 1/2 is the same as 5.5 and 1 1/4 is the same as 1.25. So you just have to subtract 1.25 from 5.5 which leaves you with 4.25 (aka 4 1/4)

5 0

3 years ago

Y - 15 =-27<br> One step equation

Serggg [28]

Add 15 to both sides
Y = -27 + 15
Y = -12

5 0

3 years ago

Read 2 more answers

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago

Determine the volume of a square pyramid that has a height of 7 inches and a base with a side

Yuri [45]

Answer:

v=58.33in^3

Step-by-step explanation:

v=a^2h/3=5^2×7/3=58.33333 in^3

3 0

3 years ago