Answer:
150 oz.
Step-by-step explanation:
There are already 150 ounces of alloy of nickel.
Of this 150 oz, 70% is pure i.e. nickel content = 150(0.7) = 105 oz
Now available is
Nickel Other metals
105 45
Let x oz of pure nickel is added.
Then new alloy will have 105+x oz nickel in total of 150+x oz.
Percentage pure = 
Simplify to get
Hence answer is 150 oz should be added.
We are given a volume of 160 fluid ounces of chemical which is added to a container that holds 120,000 gallons of water. Assuming that the chemical has the same density as water, we just need to convert 120,000 gallons to ounces.
A conversion factor is taken from literature, 1 gallon is equivalent to 128 fluid ounces. So 160 fluid ounces is only 1.25 gallons, thus occupying minimal space in the container. The employee could add more of the chemical in the container. He can actually add 15360000 fluid ounces in total.
Answer: ∠Z ≅ ∠G and XZ ≅ FG or ∠Z ≅ ∠G and XY ≅ FE are the additional information could be used to prove that ΔXYZ ≅ ΔFEG using ASA or AAS.
Step-by-step explanation:
Given: ΔXYZ and ΔEFG such that ∠X=∠F
To prove they are congruent by using ASA or AAS conruency criteria
we need only one angle and side.
1. ∠Z ≅ ∠G(angle) and XZ ≅ FG(side)
so we can apply ASA such that ΔXYZ ≅ ΔFEG.
2. ∠Z ≅ ∠G (angle)and ∠Y ≅ ∠E (angle), we need one side which is not present here.∴we can not apply ASA such that ΔXYZ ≅ ΔFEG.
3. XZ ≅ FG (side) and ZY ≅ GE (side), we need one angle which is not present here.∴we can not apply ASA such that ΔXYZ ≅ ΔFEG.
4. XY ≅ EF(side) and ZY ≅ FG(side), not possible.
5. ∠Z ≅ ∠G(angle) and XY ≅ FE(side),so we can apply ASA such that
ΔXYZ ≅ ΔFEG.
The volume of a sphere is given by:
So, we need to deduct this equation. We will walk through Calculus on the concept of a solid of revolution that is a solid figure that is obtained by rotating a plane curve around some straight line (the axis of revolution<span>) that lies on the same plane. We know from calculus that:
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Then, according to the concept of solid of revolution we are going to rotate a circumference shown in the figure, then:
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Isolationg y:
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So,
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![V=\pi \int_{a}^{b}[\sqrt{r^{2}-x^{2}}]^{2}dx](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5Cint_%7Ba%7D%5E%7Bb%7D%5B%5Csqrt%7Br%5E%7B2%7D-x%5E%7B2%7D%7D%5D%5E%7B2%7Ddx)
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being -r and r the limits of this integral.
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Solving:
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![V=\pi[r^{2}x-\frac{x^{3}}{3}]\right|_{-r}^{r}](https://tex.z-dn.net/?f=V%3D%5Cpi%5Br%5E%7B2%7Dx-%5Cfrac%7Bx%5E%7B3%7D%7D%7B3%7D%5D%5Cright%7C_%7B-r%7D%5E%7Br%7D)
Finally:
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