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3 years ago

Help will give Brainlest only true answers only please dont guess

Mathematics

1 answer:

attashe74 [19]3 years ago

7 0

Answer: 75%

Step-by-step explanation:

35% of 5 liters is 1.75 liters

1.75 liters + 8 liters = 9.75 liters (only juice)

(5 liters + 8 liters = 13 liters) (the whole thing)

So, 9.75/13 is 75%

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Suppose that 600 yards of fencing material are available to fence in

postnew [5]

Answer:

Length = 150 yards

Width = 100 yards

Step-by-step explanation:

We want 600 yards of fencing that will result in the largest 2 fenced corrals, sharing a common border.

It will take the shape of a rectangle, with a dividing fence down the center.

Let W and L, Width and Length of the larger enclosure.

See attachment.

W= Area of the larger enclosure.

The perimeter is 2W + 2L.

The dividing fence is 1W

We know that we only have 600 yards of fence, so:

2W + 2L + 1W = 600 yards

Area = W x L

---

3W + 2L = 600 (yards)

2L = 600 -3W

L = (600-3W)/2

L = 300 -(3/2)W

---

Use this expression in the Area calculation:

Area = W x L

Area = W x (300 -(3/2)W)

Area = 300W -(3/2)W^2)

To find the maximum area, take the first derivative and set to zero to find the value of W that results in the greatest area.

Area' = 300 -2(3/2)W)

0 = 300 - 3W

3W = 300

W = 100 yards

Since 3W + 2L = 600

L = (600 - 3W)/2

L = (600 - 3(100))/2

L = 150 yards

Area = 150*100 = 15,000 yards^2

8 0

3 years ago

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$