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3 years ago

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Fencing a feed lot peter plans to fence off a square feed lot and then cross-fence to divide the feed lot into four smaller squa

re feed lots. if he uses 480 ft of fencing, then how much area will be fenced in?

1 answer:

velikii [3]3 years ago

8 0

<span>6x=480 ft
x=80 ft.
Area = x^2 = 80^2 = 6400 sq. ft.</span>

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A truck is being filled with cube-shaped packages that have side lengths of 1/4 foot. The part of the truck that is being filled

n200080 [17]

Answer:

24000 pieces.

Step-by-step explanation:

Given:

Side lengths of cube = $\frac{1}{4} \ foot$

The part of the truck that is being filled is in the shape of a rectangular prism with dimensions of 8 ft x 6 1/4 ft x 7 1/2 ft.

Question asked:

What is the greatest number of packages that can fit in the truck?

Solution:

First of all we will find volume of cube, then volume of rectangular prism and then simply divide the volume of prism by volume of cube to find the greatest number of packages that can fit in the truck.

$Volume\ of\ cube =a^{3}$

$=\frac{1}{4} \times\frac{1}{4}\times \frac{1}{4} =\frac{1}{64} \ cubic \ foot$

Length = 8 foot, Breadth = $6\frac{1}{4} =\frac{25}{4} \ foot$ , Height = $7\frac{1}{2} =\frac{15}{2} \ foot$

$Volume\ of\ rectangular\ prism =length\times breadth\times height$

$=8\times\frac{25}{4} \times\frac{15}{2} \\=\frac{3000}{8} =375\ cubic\ foot$

The greatest number of packages that can fit in the truck = Volume of prism divided by volume of cube

The greatest number of packages that can fit in the truck = $\frac{375}{\frac{1}{64} } =375\times64=24000\ pieces\ of\ cube$

Thus, the greatest number of packages that can fit in the truck is 24000 pieces.

7 0

2 years ago

Find the sum of the first 6 terms of the sequence: –3, 1, 5, 9, . . .A) 45B) 42C) 48D) 25

Leni [432]

SOLUTION

From the sequence give

–3, 1, 5, 9, . . .

The first term, a = -3

The common difference, d = 4 (gotten by adding 4 to the next term).

The number of terms required n = 6.

Formula for sum of an arithmetic sequence is given by

$S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack$

Substituting these values into the equation above we have

$\begin{gathered} S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack \\ S_6=\frac{6}{2}\lbrack2\times-3+(6-1)4\rbrack \\ S_6=3\lbrack-6+(5)4\rbrack \\ S_6=3\lbrack-6+20\rbrack \\ S_6=3\lbrack14\rbrack \\ S_6=42 \end{gathered}$

Hence, the answer is 42, option B

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kondor19780726 [428]

Answer: i think your missing the other part of the question

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2 years ago

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Yuki888 [10]

2(x + 3) = 18
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2 years ago

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