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3 years ago

An Orchestra has twice as many woodwind instruments as brass instruments. There are a total of 150 brass and wood instruments.

Mathematics

1 answer:

dlinn [17]3 years ago

7 0

Answer:

Brass + Brass = Woodwind

Brass + Woodwind = 150

Brass = 50, Woodwind = 100

Step-by-step explanation:

Hi there!

The two addition equations would be:

Brass + Brass = Woodwind

(As there are twice as many woodwind as brass)

AND

Brass + Woodwind = 150

(As there are 150 of those instruments in total)

These two equations fully describe this situation, and you can use them to solve the problem using substitution as well.

Since Woodwind = Brass + Brass, then:

Brass + Woodwind =150

Can also be written as:

Brass + (Brass+Brass) = 150

Now we can solve by combining all the like terms which gives us:

3*Brass = 150

Brass = 150/3

Brass = 50 meaning that Woodwind = 100, as 50+50=100.

Hope this helped!

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3241004551 [841]

Answer:

$x=\frac{179}{3}$

Step-by-step explanation:

We are solving for $x$ in the equation:

$3x+56=235$

First, isolate the variable term by subtracting $56$ from both sides of the equation:

$3x=179$

Now, divide both sides of the equation by the coefficient of $x$ :

$x=\frac{179}{3}$

This solution for $x$ , as a decimal, would be non-terminating. If you divided $179$ into $3$ , you would get the non-terminating decimal of:

$59.66666...$

Therefore, our solution is:

$x=\frac{179}{3}$

We can check our solution by substituting $\frac{179}{3}$ for $x$ in the initial equation:

$3x+56=235$

Substitute:

$3(\frac{179}{3} )+56=235$

Simplify $3(\frac{179}{3} )$ :

$179+56=235$

Add:

$235=235$

Since both sides of the equation are equal, our solution is correct!

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3 years ago

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grin007 [14]

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I think the answer is 3.14

Step-by-step explanation:

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A teacher wants to buy some oranges for her twenty-four students to share for a snack. Each orange will be cut into eight slices

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The answer is 12. 1 orange can feed 2 kids so 1 orange=2 kids, 2 oranges=4 kids so on and so on then you get to 12 oranges=24 kids. hope this helps

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4 years ago

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Answer:

1/343

Step-by-step explanation:

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3 years ago

A. Find the common factors of 12 and 20.

Alchen [17]

Answer:

a. Common factors of 12 and 20 = 2, 2

2. Highest common factor of 12 and 20 = 4

3 packs of workbooks and 5 packs of text books to be purchased.

Step-by-step explanation:

a.) To find the common factors of 12 and 20:

First of all, let us find the factors of 12 and 20

12 = 2 $\times$ 2 $\times$ 3

20 = 2 $\times$ 2 $\times$ 5

The common factors are underlined above i.e. 2, 2.

2. The highest common factor between 12 and 20 is 2 $\times$ 2 = 4

-------------------------------------------

Workbooks are in packs of 25

Textbooks are in pack of 15

The school needs exactly same number of workbooks and textbooks.

And we have to find such least number.

This least number can simply be calculated by finding the LCM (Least Common Multiple).

LCM of two numbers p and q is the minimum number which is completely divisible by both the numbers p and q.

Let us find LCM of 25 and 15 using factorization method:

15 = 3 $\times$ 5

25 = 5 $\times$ 5

The common part is considered only once:

So LCM = 3 $\times$ 5 $\times$ 5 = 75

Hence, number of packs of workbooks to be bought = $\frac{75}{25} = 3$

number of packs of textbooks to be bought = $\frac{75}{15} = 5$

So, the answer are:

a. Common factors of 12 and 20 = 2, 2

2. Highest common factor of 12 and 20 = 4

3 packs of workbooks and 5 packs of text books to be purchased.

8 0

4 years ago