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Ivenika [448]
3 years ago
9

What can you say about the relationship between M and P? Which letter is the correct answer?

Mathematics
1 answer:
Dimas [21]3 years ago
6 0

Answer:

Choice C is correct

Step-by-step explanation:

The first step is to re-write the equations in exponential form.

The first equation can be written as;

\frac{M}{N}=10^{4} since the base is 10 and 4 the exponent.

The second equation can be written as;

\frac{P}{N}=10^{5}

The second step is to make N the subject of the formula in both equations.

Solving for N from this equation \frac{M}{N}=10^{4}, yields;

N=\frac{M}{10^{4}}

Solving for N from the second equation \frac{P}{N}=10^{5}, yields;

N=\frac{P}{10^{5}}

Therefore;

\frac{M}{10^{4}}=\frac{P}{10^{5} }\\\\P=10M

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Answer:

x = 1

Step-by-step explanation:

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2 years ago
Which of the following matrices is the solution matrix for the given system of equations? x + 5y = 11 x - y = 5
givi [52]
<h2>The required solution is x = 6 and y = 11 </h2>

Step-by-step explanation:

Given system of equations are

x+5y = 11 and x-y =5

A= \left[\begin{array}{cc}1&5\\1&-1\end{array}\right]                            X=\left[\begin{array}{c}x\\y\end{array}\right]

and          B= \left[\begin{array}{c}11\\5\end{array}\right]

∴AX=B

adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]

A= \left|\begin{array}{cc}1&5\\1&-1\end{array}\right|=-6

∴A^{-1} =\frac{adj A}{|A|}

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A^{-1} ={ \left[\begin{array}{c \c}  {{\frac{1}{6} }}&{\frac{5}{6}}\ \\  {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}

X =A^{-1}\times B

⇒\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c}  {{\frac{1}{6} }}&{\frac{5}{6}}\ \\  {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]

⇒\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c}  {6}\\  {11} \end{array}\right]}

∴ x= 6 and y = 11

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You get, x<-4.

From this, we can see that it should be an open dot at -4, and pointing to the left.

So, the correct number line is A.

7 0
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