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3 years ago

I’ll mark you brainlist show work

Mathematics

1 answer:

masya89 [10]3 years ago

3 0

Answer:

whats a slope?

Step-by-step explanation:

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What is the area of a circle who's radius is 4 ft?

masya89 [10]

Answer:

50 .27 rounded up

Step-by-step explanation:

area = pi * R squared

so it's 50 .27

6 0

3 years ago

Read 2 more answers

Caculate The Median 19, 21, 7, 22, 33, 24 help- ~jaii

emmasim [6.3K]

Order all the numbers:

7, 19, 21, 22, 24, 33

The 2 middle numbers are 21 and 22

Add both numbers and divide by 2:

21 + 22 = 43

43 / 2 = 21.5

The median is 21.5

Hope this helped☺☺

7 0

3 years ago

Positive, negative, zero or undefined?

Liula [17]

Answer:

Negative.

Step-by-step explanation:

Hope this helps.

4 0

3 years ago

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The original cost of a cell phone is x dollars. The phone is on sale for 35% off. What would be the simplified expression for th

Nonamiya [84]

X · 0.65 = y or x-(x · 0.35)= y y=sales cost

5 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