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jolli1 [7]
3 years ago
12

Please remove all perfect squares from the answer \sqrt{48b^7}

Mathematics
1 answer:
jolli1 [7]3 years ago
5 0

\sqrt{48b^7} =\sqrt{16*3*b^2*b^2*b^2*b}=4b^3\sqrt{3b}

P.S. Hello from Russia

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A chemist has two alloys, one of which is 15% gold and 20% lead and the other which is 30% gold and 50% lead. How many grams of
vlabodo [156]

Answer: 490 grams of the first alloy should be used.

30 grams of the second alloy should be used.

Step-by-step explanation:

Let x represent the weight of the first alloy in grams that should be used.

Let y represent the weight of the second alloy in grams that should be used.

A chemist has two alloys, one of which is 15% gold and 20% lead. This means that the amount of gold and lead in the first alloy is

0.15x and 0.2x

The second alloy contains 30% gold and 50% lead. This means that the amount of gold and lead in the second alloy is

0.3y and 0.5y

If the alloy to be made contains 82.5 g of gold, it means that

0.15x + 0.3y = 82.5 - - - - - - - - - - - -1

The second alloy would also contain 113 g of lead. This means that

0.2x + 0.5y = 113 - - - - - - - - - - - - -2

Multiplying equation 1 by 0.2 and equation 2 by 0.15, it becomes

0.03x + 0.06y = 16.5

0.03x + 0.075y = 16.95

Subtracting, it becomes

- 0.015y = - 0.45

y = - 0.45/- 0.015

y = 30

Substituting y = 30 into equation 1, it becomes

0.15x + 0.3 × 30 = 82.5

0.15x + 9 = 82.5

0.15x = 82.5 - 9 = 73.5

x = 73.5/0.15

x = 490

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3 years ago
Mrs. Lane took a survey of the types of pants her students were wearing. She collected the data: jeans 14, shorts 9, and capris
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36%
14+9+2=25
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2 years ago
Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp
Sedaia [141]

It looks like you're given

<em>F'(x)</em> = 3<em>x</em>² + 7

and

<em>F</em> (0) = 5

and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, <em>F</em> (0) = 5.

For the remaining <em>b</em>, you can solve for <em>F(x)</em> exactly by using the fundamental theorem of calculus:

F(x)=F(0)+\displaystyle\int_0^x F'(t)\,\mathrm dt

F(x)=5+\displaystyle\int_0^x(3t^2+7)\,\mathrm dt

F(x)=5+(t^3+7t)\bigg|_0^x

F(x)=5+x^3+7x

Then <em>F</em> (0.1) = 5.701, <em>F</em> (0.2) = 6.408, <em>F</em> (0.5) = 8.625, and <em>F</em> (2.0) = 27.

On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is

<em>F(x)</em> ≈ <em>L(x)</em> = <em>F</em> (0) + <em>F'</em> (0) (<em>x</em> - 0) = 5 + 7<em>x</em>

Then <em>F</em> (0) = 5, <em>F</em> (0.1) ≈ 5.7, <em>F</em> (0.2) ≈ 6.4, <em>F</em> (0.5) ≈ 8.5, and <em>F</em> (2.0) ≈ 19. Notice how the error gets larger the further away <em>b </em>gets from 0.

A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.

Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have

<em>x</em>₁ = <em>x</em>₀ + 0.1 = 0.1

<em>y</em>₁ = <em>y</em>₀ + <em>F'(x</em>₀<em>)</em> (<em>x</em>₁ - <em>x</em>₀) = 5 + 7 (0.1 - 0)   →   <em>F</em> (0.1) ≈ 5.7

<em>x</em>₂ = <em>x</em>₁ + 0.1 = 0.2

<em>y</em>₂ = <em>y</em>₁ + <em>F'(x</em>₁<em>)</em> (<em>x</em>₂ - <em>x</em>₁) = 5.7 + 7.03 (0.2 - 0.1)   →   <em>F</em> (0.2) ≈ 6.403

<em>x</em>₃ = <em>x</em>₂ + 0.3 = 0.5

<em>y</em>₃ = <em>y</em>₂ + <em>F'(x</em>₂<em>)</em> (<em>x</em>₃ - <em>x</em>₂) = 6.403 + 7.12 (0.5 - 0.2)   →   <em>F</em> (0.5) ≈ 8.539

<em>x</em>₄ = <em>x</em>₃ + 1.5 = 2.0

<em>y</em>₄ = <em>y</em>₃ + <em>F'(x</em>₃<em>)</em> (<em>x</em>₄ - <em>x</em>₃) = 8.539 + 7.75 (2.0 - 0.5)   →   <em>F</em> (2.0) ≈ 20.164

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2 years ago
Multiply to convert pounds to ounces. Write your answer as a fraction.
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