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Naddik [55]
3 years ago
6

What is the next term of 500, 100, 20

Mathematics
2 answers:
Fudgin [204]3 years ago
7 0

Answer: i believe it would be 5 if not i’m sorry.

Step-by-step explanation:

FinnZ [79.3K]3 years ago
6 0
I think maybe it is 4

4 x 5 = 20

20 x 5 = 100

100 x 5 = 500
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5.6 x N= 2.24 N=? Please help
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Answer:

N = -3.36

Step-by-step explanation:

5.6 x N= 2.24

-5.6          -5.6

N= -3.36

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A corner of a rectangle is cut, creating a trapezoid.
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45+90+90+x=360
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Which expression is the simplest form of -(4x + y)n + 2(x - 3y)
Marina86 [1]

Answer:

\boxed{\bold{-4nx-ny+2x-6y}}

Explanation:

Rewrite:

= \bold{-n\left(4x+y\right)+2\left(x-3y\right)}

Expand \bold{-n\left(4x+y\right): \ -4nx-ny}

= \bold{-4nx-ny+2\left(x-3y\right)}

Expand \bold{2\left(x-3y\right): \ 2x-6y}

= \bold{-4nx-ny+2x-6y}

Mordancy.

4 0
3 years ago
Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
Anit [1.1K]

Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

4 0
3 years ago
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