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Katen [24]
3 years ago
5

Someone can you please help me on number 74

Mathematics
2 answers:
SVETLANKA909090 [29]3 years ago
6 0

<u>Answer:</u>

9t^3+t^2

<u>Step-by-step explanation:</u>

We are given a figure of a polygon with mentioned side lengths and we are to find the perimeter of it.

For that, we will simply add the given side lengths and simplify them.

Perimeter of polygon = ( 4 t ^ 3 - 5 ) + ( 4 t ^ 3 - 5 ) + ( t ^ 2 + 9 ) + ( t ^ 2 + 1 2 ) + ( t ^ 3 - t ^ 2 - 1 1 )

= 4 t ^ 3 + 4 t ^ 3 + t ^ 3 + t ^ 2 - t ^ 2 + t ^ 2 - 5 - 5 + 9 - 1 1 + 1 2

Perimeter of polygon = 9t^3+t^2

lisabon 2012 [21]3 years ago
5 0

Answer:

  9t^3 +t^2

Step-by-step explanation:

The perimeter of the figure is the sum of the lengths of the sides. The side lengths are represented by the polynomials shown, so the perimeter (P) is their sum:

  P = (4t^3 -5) + (4t^3 -5) + (t^2 +9) + (t^3 -t^2 -11) + (t^2 +12)

Rearranging to group like terms:

  P = (4t^3 +4t^3 +t^3) + (t^2 -t^2 +t^2) + (-5 -5 +9 -11 +12)

  P = 9t^3 +t^2

The perimeter of the figure is represented by the polynomial 9t^3 +t^2.

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\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}

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The volume of a pyramid varies jointly as the height and the area of the base. If a pyramid has the measurements V=1144
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  • V = (1/3)Bh

Step-by-step explanation:

The given relation is ...

  V = kBh . . . . . for some base area B, height h, and constant of variation k

We are given length and width of the base so we presume it is a rectangle.

  B = l·w = 8·11 = 88 . . . . square meters

The given volume tells us the value of k:

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The value of k is about 0.33.

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Then the volume of the larger pyramid is ...

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The general relationship is ...

  V = 1/3Bh

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