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Elden [556K]
3 years ago
7

How does 8 4/3 = 9 1/3?​

Mathematics
1 answer:
irga5000 [103]3 years ago
7 0

You're question asks why the given mixed numbers are equal.

In fact, they're equal.

Lets make it easier for you to understand:

8 4/3 has a improper fraction "4/3", that fraction is over 1 whole.

You could look at 8 4/3 as:

8 + 3/3 + 1/3

The 3/3 turns into 1:

8 + 1 + 1/3

9 + 1/3

9 1/3

You would see that it turns into 9 1/3, therefore they are equal.

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What statements are correct? Check all that apply. The conditional probability formula is P(x l n m PO) The conditional probabil
Monica [59]

The first , second, and fourth options if you are on edge.

4 0
4 years ago
you have 4/6 cup of flour in your cupboard a recipe for bread calls for 1/4 cup of flour how much flour would you have left if y
andreyandreev [35.5K]
Hello,

Here is your answer:

The proper answer to your question is "5/12".

Here is how:

All you have to do is subtract 4/6 and 1/4:

4×2=8
6×2=12
-
1×3=3
4×3=12

8-3=5

Your answer is 5/12!

If you need anymore help feel free to ask me!

Hope this helps!
7 0
3 years ago
shen's mother gave him $10 to go buy groceries. he picked up a gallon oh milk for $3.49,tree pounds of oranges that cost $1.14 p
bekas [8.4K]

Answer:

He didn’t have enough money for the groceries

Step-by-step explanation:

He started out with $10

He spent $3.49 on a gallon of milk

He spent $1.14 on 3 pounds of oranges each

He spent $3.46 on cereal


The total cost would be $10.37!


8 0
3 years ago
A basketball player has a 60% chance of making a free throw what is the probabilty the player will make 3 out of 6
ozzi
It will be 10% that is half of it
7 0
3 years ago
How do you solve this?​
rewona [7]

\text{L.H.S}\\\\=\begin{vmatrix} 1 & bc& b+c\\ 1 & ca& c+a\\ 1 &ab&a+b\end{vmatrix}\\\\\\=\begin{vmatrix} 1 & bc& b+c-(a+b+c)\\ 1 & ca& c+a-(a+b+c)\\ 1 &ab&a+b-(a+b+c)\end{vmatrix}~~~~~~~~~~~~~~~;c_3\rightarrow c_3 -(a+b+c)c_1\\\\\\=\begin{vmatrix} 1 & bc& -a\\ 1 & ca& -b\\ 1 &ab&-c\end{vmatrix}\\\\\\=\dfrac 1{abc}\begin{vmatrix} a & abc& -a^2\\ b & abc& -b^2\\ c &abc&-c^2\end{vmatrix}\\\\\\=-\dfrac {abc}{abc}\begin{vmatrix} a & 1& a^2\\ b & 1& b^2\\ c &1&c^2\end{vmatrix}\\

=-1\begin{vmatrix} a & 1& a^2\\ b & 1& b^2\\ c &1&c^2\end{vmatrix}\\\\\\=-1 \times -1 \begin{vmatrix} 1 & a& a^2\\ 1 & b& b^2\\ 1 &c&c^2\end{vmatrix}\\\\\\=\begin{vmatrix} 1 & a& a^2\\ 1 & b& b^2\\ 1 &c&c^2\end{vmatrix}\\\\\\=\text{R.H.S}\\\\\text{Showed.}

3 0
2 years ago
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