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nadezda [96]
3 years ago
13

What is 51/7 by 31/9 in simplest form

Mathematics
1 answer:
konstantin123 [22]3 years ago
3 0

Answer:

25,095238095 = 25 2⁄21

Step-by-step explanation:

Just multiply each denominator and numerator straight across and convert to a mixed number, decimal, or whatever you want.

I am joyous to assist you anytime.

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Solve for this problem for N
mezya [45]

Answer:

N = 3

Step-by-step explanation:

I don't know what the whole thing above is, but just disregard that.

All it is here is cross multiplication.

So 28N = 21 * 4

Multiply...

28N = 84

And divide each side by 28

N = 3

4 0
3 years ago
Read 2 more answers
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
3 years ago
Can anybody help me with this?<br><br> Use this graph of f to find the solutions to f(x) = 0.
Brums [2.3K]
Another way to say f(x) is y, so y=0 at -2 and 10. :)
6 0
3 years ago
Solve for x. <br> please help me!!
Nady [450]

Answer:

(2)(18)x

=36x

Step-by-step explanation:

sorry if its wrong i need points and i didnt quit know what this was but sorry if this was wrong and sorry if i didnt help

7 0
3 years ago
Donnie draws four cards from a fair deck of 52. A fair deck of 52 has one King of Hearts. What is the approximate probability of
My name is Ann [436]

Answer:

B)

Step-by-step explanation:

Since Donnie drew four cards, his probibilty of drawing the one King Of Hearts

8 0
3 years ago
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