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Amiraneli [1.4K]
3 years ago
9

Write 39,005 in expanded form

Mathematics
2 answers:
vlada-n [284]3 years ago
8 0
(3 * 10,000) + ( 9* 1,000) + ( 5* 1)
8090 [49]3 years ago
5 0
Just go up by ones tes hundreds thousands and so on to get - 30,000+9000+5
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If f(x) = -3x-5 and g(x)=4x-2, find (f-g)(x)
MA_775_DIABLO [31]

f(x) -g(x) = -3x-5 -(4x-2)

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-3x-5 -4x+2

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-7x-3

Answer:

-7x-3

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3 years ago
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What is the value of x?
stiks02 [169]

Answer:

Step-by-step explanation:

Use Pythagoras Theorem

x² + 16² = 34²

x² + 256 =  1156

x² = 1156 - 256

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To get the volume, you need the area of the sides of the square.

The total surface area is 96, with 6 sides. To get the surface area of just one side, divide 96 by 6, which gets you 16. The surface area of a single square is 16. To get one side of the square, divide that by 4 because there are 4 sides.

Now, multiply 16 by 4 to get the volume.

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3 years ago
I can’t figure this one out! Someone please help
Zinaida [17]

Answer:

g= -2x+1

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Estimate the following quantities without using a calculator. Then find a more precise result, using a calculator if necessary.
zubka84 [21]

Explanation:

a. It is so easy to double a number that no approximation is necessary.

  31,000 × 200 = 3.1×10⁴ × 2×10² = 6.2×10⁶ exactly

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b. 6.2×10³ × 5.2×10⁶ ≈ 6×5×10⁹ = 3×10¹⁰ approximately

The approximation can be refined a bit by taking the ".2" into account:

  6.2×10³ × 5.2×10⁶ ≈ (6×5 + .2×(6+5))×10⁹

  = 32.2×10⁹ = 3.22×10¹⁰ approximately

Actual product: 3.224×10¹⁰.

For most purposes, the approximation is an adequate approximation, as it it within 10% of the actual value.

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c. 9×10⁶ -2.3×10⁴ ≈ 9×10⁶ approximately

A better approximation is to actually subtract an approximation of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.02)×10⁶ ≈ 8.98×10⁶ approximately

The actual value uses all of the digits of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.023)×10⁶ = 8.977×10⁶ exactly

As in part B, either approximation is adequate for most purposes, as the difference from the actual value is less than .3%.

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The accuracy required of an approximation, hence the work you expend improving accuracy, should depend on the need in the final application of the number. Often, approximations are used for budget or resource planning purposes where some "slop" is allowed or even expected.

They can also be used in engineering applications, where the error needs to be on the side of more safety (rather than less).

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3 years ago
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