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1 year ago

Simplify: −3 • −26 A. −78 B. −68 C. 68 D. 78

1 answer:

3 0

-3 x -26 = 78

meaning the answer is D.

The reason the answer is D, is because two negatives when multiplied or divided become positive. Only when multiplied or divided does this occur.

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Two triangles are congruent if

irinina [24]

Answer:

B is your answer

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Write the first five terms of the sequence in which the nth term is

ziro4ka [17]

Answer:

C) 2,6,24,120,720

Step-by-step explanation:

Here the n-th term An = $\frac{(n +2)!}{n + 2}$

A1 is the first term

A1 = (1 + 2)!/(1 + 2)

= 3!/3

A1 = 2

A2 is the second term

A2 = (2 +2)!/(2 +2)

= 4!/4

A2 = (1*2*3*4) /4

A2 = 6

A3 is the third term

A3 = (3 + 2)!/(3 +2)

A3 = 5!/5

A3 = 24

A4 is the fourth term

A4 = (4 + 2)!/(4 + 2)

A4 = 6!/6

A4 = 120

A5 is the fifth term

A5 = (5 + 2)!/(5 +2)

A5 = 7!/7

A5 = 720

Answer: C) 2,6,24,120,720

Thank you.

8 0

2 years ago

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

If m is positive and k is negative what is m-k

Andrew [12]

M+k because the subtraction sign and negative cancel out to become positive

8 0

3 years ago

Read 2 more answers

Find the angle of RMQ.

faust18 [17]

Answer:

Step-by-step explanation:

since there is a bisector that makes the two angles equal. you first need to find x then with x you can add the angles to get the biggest angle witch is RMQ

2x+1 = 4x-9

1=2x-9

10=2x

5=x

x=5

RMQ=2x+1+4x-9

RMQ=2(5)+1+4(5)-9

RMQ=10+1+20-9

RMQ=11+11

RMQ=22

RMQ=16

8 0

3 years ago