All categories

2 years ago

Order of operation 36divided4+11x8

Mathematics

1 answer:

frosja888 [35]2 years ago

5 0

The answer for 36/4 + 11 $\times$ 8 = 97.

<u>Step-by-step explanation</u>:

The order of operation is done by using BODMAS rule.

B - bracket

O - of

D - division

M - multiplication

A - addition

S - subtraction

step 1 :

36 divided 4 + 11 $\times$ 8

The division operation must be done first.
36/4 = 9

step 2 :

9 + 11 $\times$ 8

After division, multiplication should be done next.
11 $\times$ 8 = 88

step 3 :

9 + 88

Finally addition is performed to get the answer.
9 + 88 = 97

∴ The answer for 36divided4+11x8 is 97.

You might be interested in

21. please and thank you

olga2289 [7]

16 is the answer just add

8 0

2 years ago

F(x) = x2 what is g(x)?

Naily [24]

Answer:

i got the same problem

Step-by-step explanation:

3 0

2 years ago

Translate the phrase into an expression: <br>8 times the sum of 7 and y

Daniel [21]

Answer:

saw dude

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Express the system as AX = B; then solve using matrix inverses

Wittaler [7]

Answer with explanation:

1. The given equations are

3x -5 y=2

-x+2 y= 0

⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]$

$\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B$

Adj.A=Transpose of cofactor of Matrix A

$Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2$

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]$

$\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4$

7 0

3 years ago

Can Someone Plz help me Thank you will give brainliest!!

Akimi4 [234]

Answer:

idk

Step-by-step explanation:

sorry man i can't help u rite now but i can get u a tooter tho

7 0

2 years ago

Read 2 more answers