If 1m = 3.3 ft, how many cubic feet is the Kuroshio Sea tank? Volume is 9,450. Show your calculations!

Mathematics

1 answer:

Naddika [18.5K]2 years ago

3 0

Answer:

339604.65 cubic feet

Step-by-step explanation:

We are given that 1 m is equal to 3.3 ft.

1 m = 3.3 ft

This means that if we find the volume of a cube of 1 m, we have its volume as:

$1^3 = 1 m^3$

In the same way, the volume of a cube of 3.3 ft (1 m) is:

$3.3^3 = 35.937 ft^3$

This means that:

$1m^3 = 35.937 ft^3$

The Kuroshio sea tank has volume of $9450 m^3$ . Its volume in cubic feet is therefore:

$1m^3 = 35.937 ft^3\\\\9450 m^3 = 9450 * 35.937 = 339604.65 ft^3$

The volume of the tank is 339604.65 cubic feet.

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How do I solve this?

aksik [14]

Multiply the first equation by 2
4x-6y=16

Then add the equations to eliminate x
4x-6y=16
-4x+5y=-10

-y=6
y-=6

Plug the y value in
2x-3(-6)=8
2x+18=8
2x=-10
x=-5

Final answer: (-5,-6)

4 0

3 years ago

Read 2 more answers

Find the Y-coordinate of point P that lies 1/3 along segment RS, where R (-7, -2) and S (2, 4).

QveST [7]

Solution:

Given that the point P lies 1/3 along the segment RS as shown below:

To find the y coordinate of the point P, since the point P lies on 1/3 along the segment RS, we have

$\begin{gathered} RP:PS \\ \Rightarrow\frac{1}{3}:\frac{2}{3} \\ thus,\text{ we have} \\ 1:2 \end{gathered}$

Using the section formula expressed as

$[\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}]$

In this case,

$\begin{gathered} m=1 \\ n=2 \end{gathered}$

where

$\begin{gathered} x_1=-7 \\ y_1=-2 \\ x_2=2 \\ y_2=4 \end{gathered}$

Thus, by substitution, we have

$\begin{gathered} [\frac{1(2)+2(-7)}{1+2},\frac{1(4)+2(-2)}{1+2}] \\ \Rightarrow[\frac{2-14}{3},\frac{4-4}{3}] \\ =[-4,\text{ 0\rbrack} \end{gathered}$

Hence, the y-coordinate of the point P is