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weqwewe [10]
3 years ago
6

THIS IS VERY IMPORTANT!!!!!!!!

Mathematics
1 answer:
saul85 [17]3 years ago
6 0
I’m pretty sure it’s Abraham Lincoln, not sure though
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Arlene had 1/3 as many goldfish as Isabelle. Isabelle has 5 times as many goldfish as Bruce. If Bruce had 18 goldfish, how many
kvv77 [185]

Answer: 30 goldfish

Step-by-step explanation: Bruce has 18 goldfish. If Isabelle has 5 times as many goldfish as Bruce she will have 18*5=90 goldfish. If Arlene has 1/3 times as many goldfish as Isabelle she will have 90/3=30 goldfish.

7 0
2 years ago
Math Math Math Math help help help
Alchen [17]

Answer: 5a: 5/8

Step-by-step explanation: sorry, thats the only one i can read `:)

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2 years ago
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What percent of 38 is 12
mars1129 [50]

Answer:

31%

Step-by-step explanation:

hope this helps

8 0
2 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}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
2 years ago
80 POINTS!!! WILL VENMO $5
Hunter-Best [27]

Answer:

WAKA WAKA AFRICAAAA AFRIKA BAM BAM MAMAMAMAMAMAMAMAMAM.A.

So did I get them 5 bucks?

4 0
2 years ago
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