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NeX [460]
2 years ago
13

2

Mathematics
1 answer:
mestny [16]2 years ago
3 0
The answer for abcd is a whiff of this

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3 our of 5 in two equivalent forms
sammy [17]
Hello!

3 out of 5 looks like this as a fraction:

3/5

We cannot divide this further as both the numerator and the denominator are odd numbers, so we must multiply to get an equivalent fraction:

3/5 = 6/10 = 60/100

3/5 is also equivalent to 0.2 as a decimal or 20% as a percentage.
7 0
4 years ago
The Bowen family's pancake recipe uses 2 teaspoons of baking powder for every
Veseljchak [2.6K]

Answer:

6 teaspoon of salt

Step-by-step explanation:

All you need to do is find how many teaspoon of salt they have in total which in this case they have 1 and there are 3  1/3s  in 1 teaspoon then you just have to do 3 times 2 and you get 6teaspoon of salt.

Hope this helps!

5 0
3 years ago
alph and his brother are at a carnival. They separate from each other at the Ferris wheel at 1:00 PM, and they agree that they w
Blababa [14]

Answer: 3pm

Step-by-step explanation:

It will take them 120 minutes to meet back up together

24x5=120

15x8=120

6 0
3 years ago
Which expression is not equivalent to 8 · x · 2 · y?<br><br> 8 x + 2 y<br> 16 xy<br> 16 yx
andriy [413]

The answer is...

8x+2y

4 0
3 years ago
Read 2 more answers
Consider the matrix A =(1 1 1 3 4 3 3 3 4) Find the determinant |A| and the inverse matrix A^-1.
solong [7]

Answer:

A)\,\,det(A)=1

B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]

Step-by-step explanation:

det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|

Expanding with first row

det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1

To find inverse we first find cofactor matrix

C_{1,1}=(-1)^{1+1}\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|=7\\\\C_{1,2}=(-1)^{1+2}\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|=-3\\\\C_{1,3}=(-1)^{1+3}\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|=-3\\\\C_{2,1}=(-1)^{2+1}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=-1\\\\C_{2,2}=(-1)^{2+2}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\C_{2,3}=(-1)^{2+3}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\

C_{3,1}=(-1)^{3+1}\left\Big|\begin{array}{cc}1&1\\4&3\end{array}\right\Big|=-1\\\\C_{3,2}=(-1)^{3+2}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\\\C_{3,3}=(-1)^{3+3}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\

Cofactor matrix is

C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]

4 0
3 years ago
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