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Lerok [7]
3 years ago
13

334 divided by 12 what is the correct answer?

Mathematics
1 answer:
inessss [21]3 years ago
8 0
I got 167/6
Hope it helps
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If one side of a square is 16 how long is the diagonal?​
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Answer:

d≈22.63

Step-by-step explanation:

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4 years ago
the eatery restaurant has 200 tables on a recent evening there were reservations for one tenth of tables how many tables were re
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200 * 1/10 = 20
20 tables were reserved
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Mai is given these two systems of linear equations to solve:
Sholpan [36]

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Step-by-step explanation:

System 1 has no solution, and System 2 has infinitely many solutions. Wait, YOUR IN HIGH SCHOOL?! IM IN 8TH GRADE AND IM DOING THIS??!! well, idk how to solve this, but these are the answers.

6 0
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A rectangular Corn Hole area at the Recreation Center has a width of 5 feet and a length of 10 feet. If
lbvjy [14]

Answer:

2ft

Step-by-step explanation:

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3 0
3 years ago
I dont get it plzzz help me
cricket20 [7]

Given Equation: \frac{1}{3}h-4(\frac{2}{3}h-3)=\frac{2}{3}h-6

Let us remove parenthesis first.

In order to remove parenthesis, we need to distribute -4 over parenthesis (2/3 h -3).

Distributing -4 over (2/3 h -3), we get,

-4*2/3 h - 4*-3 = -8/3 h + 12.

Substituting this value in original equation, we get

\frac{1}{3}h-\frac{8}{3}h+12 =\frac{2}{3}h-6

Now, in order to make the equation easy to solve, we always remove fractions. In order to remove fraction, we need to find the lowest common denominator of all terms(lcd).

The fraction terms has 3 in denominators, so we could multiply each and every term by 3 to remove 3's from denominators.

Multiplying each equation by 3, we get,

3*\frac{1}{3}h-3*\frac{8}{3}h+3*12 =3*\frac{2}{3}h-3*6

On simplfying this step, we get

1h -8h+36=2h-18

Combining like terms on left side, we get

-7h +36 = 2h -18.

Subtracting 36 from both sides, we get

-7h +36 -36= 2h -18-36.

-7h=2h-54

Subtracting 2h from both sides, we get

-7h-2h=2h-54-2h

-9h = -54

Dividing both sides by -9.

-9h/-9 = -54/-9

h = 6.

Therefore, final answer is h=6.


5 0
4 years ago
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