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mars1129 [50]
3 years ago
13

What is the fraction 3/2 squared in fraction form??

Mathematics
2 answers:
ehidna [41]3 years ago
4 0
You basically just take 3^2 and 2^2 to get 9/4
9/4 is the answer

Anuta_ua [19.1K]3 years ago
4 0

Answer:

The required fraction is 9/4.

Step-by-step explanation:

The given expression is

(\frac{3}{2})^2

Power of quotient property of exponent: If a and b and are non zero real number, m and n are any integers, then

(\frac{a}{b})^m=\frac{a^m}{b^m}

Using quotient property of exponent, we get

(\frac{3}{2})^2=\frac{3^2}{2^2}

(\frac{3}{2})^2=\frac{9}{4}

Therefore the required fraction is 9/4.

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DOES ANYONE KNOW HOW TO DO THIS???
Anna11 [10]

Answer:

Cost of a pound of chocolate chips: $3.5

Cost of a pound of walnuts: $1.25

Step-by-step explanation:

x - cost of a pound of chocolate chips

y - cost of a pound of walnuts

We create two equations based on the information we have:

3x+2y=13

8x+4y=33

The whole point of these problems os to get rid of x or y. In this question, we can do this by multiplying both sides of the first equation by 2, and then subtracting it from the second equation:

8x+4y=33

6x+4y=26

2x=7

x=3.5

Then we change x for 3.5 in the first equation:

3×3.5+2y=13

10.5+2y=13

2y=2.5

y=1.25

Hope this helps!

4 0
3 years ago
Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

7 0
4 years ago
How to solve X - 1v+ 4 + 7y - 3
Nostrana [21]
The first step to solving this is to remember our mathematical rules. They state that when the term has a coefficient of -1,, the number doesn't have to be written but the sign needs to remain. This will change the expression to the following:
x - v + 4 + 7y - 3
Now subtract the numbers 4 and 3 from each other.
x - v + 1 + 7y 
Since this expression cannot be simplified any further,, the correct answer to your question would be x - v + 1 + 7y.
Let me know if you have any further questions.
:)
3 0
3 years ago
Kalvin tosses a coin five days in a row and gets tails every time. Do you think there is something wrong with the coin? How can
riadik2000 [5.3K]
I don’t think there is anything wrong with the coming there is just a 50% chance that It would land on tails and there’s a 50% chance that it will land on heads ‍♀️☺️
7 0
3 years ago
3x + 6y = 14<br>2x + 8y = 2​
solmaris [256]

Step-by-step explanation:

x=14-6y/3

substituting x=14-6y/3 into eqn 2

2(14-6y/3)+8y=2

multiplying through by 3

2(14-6y/3)×3 +8y×3=2×3

2(14-6y)+24y=6

28-12y+24y=6

28-12y=6

28-6=12y

22=12y

22/12=y

finding x

2x+8(22/12)=2

2x+44/3=2

2x=2-44/3

x=(-38/3)/2

×=-38/6

7 0
2 years ago
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