Active forum topics
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
- Sight Distance of Vertical Parabolic Curve
New forum topics
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Integration of 4x^2/csc^3x√sinxcosx dx
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Newton's Law of Cooling
- Law of cosines
- Can you help me po to solve this?
- Eliminate the Arbitrary Constants
Recent comments
- Yes.1 week 1 day ago
- Sir what if we want to find…1 week 1 day ago
- Hello po! Question lang po…3 weeks 5 days ago
- 400000=120[14π(D2−10000)]
(…2 months ago - Use integration by parts for…2 months 4 weeks ago
- need answer2 months 4 weeks ago
- Yes you are absolutely right…3 months ago
- I think what is ask is the…3 months ago
- $\cos \theta = \dfrac{2}{…3 months ago
- Why did you use (1/SQ root 5…3 months ago
To get the volume of the pile
To get the volume of the pile of ore, which looks like this:
We need to know the formula for the volume of the figure above. The formula applicable would be:
$$V = Bh$$
where $V$ is volume of the figure, $B$ is the area of base and $h$ is the height of the figure.
We notice that we don't know the area of the end bases of this pile of ore. We need to get it by using the formula to get the area of the triangle.
Getting the other sides of the pile of ore:
$$\cos 45^o = \frac{x}{60}$$ $$x = 30\sqrt{2} \space ft$$
Now getting the area of the ends of this pile of ore:
$$Area = \frac{1}{2}bh$$ $$Area = \frac{1}{2}(30\sqrt{2} \space feet)(30\sqrt{2} \space feet)$$ $$Area = 900 \space ft^2$$
Now we can get the volume of this pile of ore:
$$V= Bh$$ $$V = (900 \space ft^2)(500 \space ft)$$ $$V = 450000 \space ft^3$$
Therefore the volume of your pile of ore is $\color{green}{450000 \space ft^3}$
Alternate solutions are encouraged...heheheheh