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tekilochka [14]
3 years ago
5

How do I solve this

Mathematics
1 answer:
Flauer [41]3 years ago
3 0
Volume of Cylinder  V₁ =πR². H

Volume of a cone  V₂ = (1/3).πR².h

We know that H=4h (given) &  R, the radius =75 &The total Volume = V₁+V₂,
We also know that the hopper has one cylinder & 2 equal cones
Hence
πR². H + 2 [(1/3).πR².h] =343,000 ft³
π(75)².H + 2[(1/3).π(75)².h] =343,000
But H = 4h, then
π(75)².4h + 2[(1/3).π(75)².h] =343,000. Solve for h
26250.π.h = 81,466.8.h = 343,000

& h=343,000/81,466.8, ==> h=4.16 ft & H =16.6 FT



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Answer and explanation:

  - 5 = b - 2.3    add 2.3 to both sides

+2.3        +2.3

- 2.7 = b

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LuckyWell [14K]

Answer:568,000

Step-by-step explanation:

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Evaluate the expression
larisa86 [58]

Answer:

12-[20-2(6^2\div3\times2^2)]=88

Step-by-step explanation:

So we have the expression:

12-[20-2(6^2\div3\times2^2)]

Recall the order of operations or PEMDAS:

P: Operations within parentheses must be done first. On a side note, do parentheses before brackets.

E: Within the parentheses, if exponents are present, do them before all other operations.

M/D: Multiplication and division next, whichever comes first.

A/S: Addition and subtraction next, whichever comes first.

(Note: This is how the order of operations is traditionally taught and how it was to me. If this is different for you, I do apologize. However, the answer should be the same.)

Thus, we should do the operations inside the parentheses first. Therefore:

12-[20-2(6^2\div3\times2^2)]

The parentheses is:

(6^2\div3\times2^2)

Square the 6 and the 4:

(36\div3\times4)

Do the operations from left to right. 36 divided by 3 is 12. 12 times 4 is 48:

(36\div3\times4)\\=(12\times4)\\=48

Therefore, the original equation is now:

12-[20-2(6^2\div3\times2^2)]\\=12- [20-2(48)]

Multiply with the brackets:

=12-[20-96]

Subtract with the brackets:

=12-[-76]

Two negatives make a positive. Add:

=12+76=88

Therefore:

12-[20-2(6^2\div3\times2^2)]=88

3 0
3 years ago
What is the surface area of the pyramid whose net is shown here?
Lunna [17]
Urface area  = area of the square + area of the 4 triangles

= 6*6 + 4(1/2*6*6)  

= 36 + 4 * 18  = 108 sq cms
4 0
3 years ago
If lim x-> infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b
MAXImum [283]

We have

\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}

So

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0

and this happens if <em>a</em> = 1 and <em>b</em> = -1.

To confirm, we have

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0

as required.

3 0
3 years ago
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