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iVinArrow [24]
3 years ago
11

My cousin goes to a strange schools. It's called Geometric Middle School. My cousin's math project is to determine how many tile

s will be needed to tile the whole floor if the cafeteria is to have sides that are 12 feet long. He sent me the information and he wants my help. What is the answer?
A hexagon that is one foot long, uses 6 tiles
A hexagon with sides two feet long uses 24 tiles
Mathematics
1 answer:
Anon25 [30]3 years ago
4 0
Area of trapezoid  = (1/2) (b1 + b2) h
                             = (1/2) x  (12 + 24) x 6 sqrt 3
                             =   108 sqrt 3

area of Hexagon = 2 x Area of trapezoid
                            = 2 x 108 sqrt 3
                            = 216 sqrt 3

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Answer:

9u^3 + 6u^2 - 7u + 6

Step-by-step explanation:

7 0
3 years ago
A bag contains 1 kilogram of potatoes Each potato has a mass of approximately 3 ounces. a) Determine the approximate number of p
fgiga [73]

1 ounce = 28 grams (aprox)

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1000 grams / 28 grams = 35.7142857143

You cant have 35.7142857143 patatoes and so we round this to 35 potateos.

= 35

⭐ Answered by Foxzy0⭐

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4 0
2 years ago
A cylinder has a diameter of 10 centimeters and a height of 15 centimeters. What is the surface area of the cylinder to the near
adelina 88 [10]

The surface area of the cylinder to the nearest tenth is:  628.3

The surface area without rounding is:  628.32

7 0
3 years ago
You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%
bezimeni [28]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}

\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}

6 0
3 years ago
Which expressions are equivalent to -f-5(2f-3)
gizmo_the_mogwai [7]

Answer:

-11f +15

Step-by-step explanation:

-f-5(2f-3)

Distribute

-f -10f +15

Combine like terms

-11f +15

3 0
3 years ago
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