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Leviafan [203]
2 years ago
12

A giant scoop, operated by a crane, is in the shape of a hemisphere of radius 21 inches. The scoop is used to transfer molten st

eel into a cylindrical storage tank with diameter 28 inches. If the scoop is filled to the top, what will be the height of the molten steel in the storage tank?
Write the answer as a decimal, rounded to the nearest tenth.
Mathematics
1 answer:
eduard2 years ago
3 0

Answer:

31.5 in

Step-by-step explanation:

  • Volume of a hemisphere = (2/3)\pir³
    (where r is the radius)
  • Volume of a cylinder = \pir²h
    (where r is the radius and h is the height)
  • radius r = (1/2) diameter

First, find the volume of the scoop using the volume of a hemisphere formula with r = 21:

Volume = (2/3)\pi x 21³ = 6174\pi in³

Now equate the found volume of the scoop to the equation of the volume of a cylinder with r = 14, and solve for h:

                                            \pi14²h = 6174\pi

                                            196\pih = 6174\pi

 Divide both sides by \pi:        196h = 6174

Divide both sides by 196:          h = 31.5

Therefore, the height of the molten steel in the storage tank is 31.5 in

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Dividing by a fraction is equivalent to multiply by its reciprocal, then:

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Now, we need to express the quadratic polynomials using their roots, as follows:

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Applying the quadratic formula to the first polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}

Applying the quadratic formula to the second polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the third polynomial:

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Applying the quadratic formula to the fourth polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}

Substituting into the rational expression and simplifying:

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8 0
1 year ago
Identify the monomial function described as odd or even, and indicate whether a is positive or negative.
Igoryamba

Answer:

Case 1: As x > 0 increases, f(x) increases.  As x < 0 decreases, f(x) decreases.

It is an 'odd' function with 'positive' a.

Case 2: As x > 0 increases, f(x) decreases.  As x < 0 decreases, f(x) decreases.

It is an 'even' function with 'negative' a.

Case 3: As x > 0 increases, f(x) increases.  As x < 0 decreases, f(x) increases.

It is an 'even' function with 'positive' a.

Case 4: As x > 0 increases, f(x) decreases.  As x < 0 decreases, f(x) increases.

It is an 'odd' function with 'negative' a.

Step-by-step explanation:

Let us consider a monomial function:

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Case 1:

As x > 0 increases, f(x) increases.  As x < 0 decreases, f(x) decreases.

This happens only if a is 'positive' and n is 'odd'.  So, it is an 'odd' function with 'positive' a.

Case 2:

As x > 0 increases, f(x) decreases.  As x < 0 decreases, f(x) decreases.

This happens only if a is 'negative' and n is 'even'. So, it is an 'even' function with 'negative' a.

Case 3:

As x > 0 increases, f(x) increases.  As x < 0 decreases, f(x) increases.

This happens only if a is 'positive' and n is 'even'. So, it is an 'even' function with 'positive' a.

Case 4:

As x > 0 increases, f(x) decreases.  As x < 0 decreases, f(x) increases.

This happens only if a is 'negative' and n is 'odd'. So, it is an 'odd' function with 'negative' a.

Keywords: monomial function, odd function, even function

Learn more about monomial function from brainly.com/question/8973176

#learnwithBrainly

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3 years ago
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Answer:

Step-by-step explanation:

Here you go mate

Step 1

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Step 2

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3x(1.05)

Step 3

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Answer

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Hope this helps

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