Need help figuring out the measurement here

Mathematics

1 answer:

tresset_1 [31]3 years ago

7 0

Answer:

Step-by-step explanation:

4m-1m=3m

Or you can just count

Hope this helps! :)

You might be interested in

Factor the expression 20g3 + 24g2 – 15g – 18

abruzzese [7]

First, put parenthesis around the first two numbers and the last two numbers.
(20g³+24g²) (-15g-18)
Then, take out the greatest common factor of both parenthesis.
4g²(5g+6)-3(5g+6)
You then separate the numbers outside the parenthesis and the numbers in the parenthesis.
(5g+6) (4g²-3)
Then you simplify the second set of numbers. Since the set of numbers can't be simplified, you would leave this problem as it is. I hope this makes sense.

3 0

3 years ago

Salama makes fruit baskets with oranges and bananas.

vovangra [49]

The ratio for problem is:

oranges : bananas = 9 : 6 or oranges : bananas = 6 : 4

Solution:

Given that, Salama makes fruit baskets with oranges and bananas

She uses between 8 and 16 oranges and bananas in a ratio of 3 oranges to 2 bananas

Given ratio is:

Ratio of orange to bananas = 3 : 2

They add up to 5 fruits, so some possible values are 5, 10, 15 , 20, 25, etc

But we need a value between 8 and 16 oranges and bananas

So we can choose 15 or 10

Now let us choose 15 and solve

Let the oranges be 3x

Let the bananas be 2x

Therefore,

3x + 2x = 15

5x = 15

x = 3

Thus oranges = 3x = 3(3) = 9

Bananas = 2x = 2(3) = 6

Thus the ratio for problem is:

oranges : bananas = 9 : 6

Similarly solve for 10

3x + 2x = 10

5x = 10

x = 2

Thus oranges = 3x = 3(2) = 6

Bananas = 2x = 2(2) = 4

Thus the ratio for problem is:

oranges : bananas = 6 : 4

8 0

3 years ago

What is the difference in simplest form? <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bn%5E2%2B3n%2B2%7D%7Bn%5E2%2B6n%2B8%7D-%5

Paha777 [63]

$\bf \cfrac{n^2+3n+2}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4}
\\\\\\
\textit{so our LCD will just be (n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~2n^2-4n}{(n+4)(n+2)}
\implies 
\cfrac{-n^2-n+2}{(n+4)(n+2)}
\\\\\\
\cfrac{-(n^2+n-2)}{(n+4)(n+2)}\implies \cfrac{-\underline{(n+2)}(n-1)}{(n+4)\underline{(n+2)}}\implies \cfrac{-(n-1)}{n+4}\implies \cfrac{1+n}{n+4}$