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2 years ago

In a translation, what math process is used? WILL GIVE BRAINILEST!!!

Mathematics

2 answers:

larisa86 [58]2 years ago

5 0

The answer is b hope this helps

jekas [21]2 years ago

3 0

The answer is B multiplication That’s correct who is the

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Rewrite using exponential notation \root(6)((4\pi )^(3))

Nezavi [6.7K]

$~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] ~\dotfill\\\\ \sqrt[6]{(4\pi )^3}\implies (4\pi )^{\frac{3}{6}}\implies (4\pi )^{\frac{1}{2}}$

5 0

1 year ago

What is the equation of the line that passes through the point (-2,0) and has a slope of -2

Tcecarenko [31]

Answer:

y = -2x -4

Step-by-step explanation:

The equation of a line is y = mx + b, where m is the slope of the line, and b is the y-intercept. We are told that the slope of the line is -2, so we can start with

y = mx + b

y = -2x + b

We can figure out the value of b by plugging in the coordinates of the point given (-2, 0)

0 = -2(-2) + b

0 = 4 + b

Subtract 4 from both sides of the equation

-4 = b

Then plug that value of b back into the equation above:

y = -2x + (-4) or y = -2x -4

6 0

3 years ago

Read 2 more answers

Help me out please :)

Veseljchak [2.6K]

Answer: -6 is the answer

Step-by-step explanation:

4 0

3 years ago

Find BC round to the nearest tenth

slega [8]

Answer:

6.7

Step-by-step explanation:

Here, we can apply cosine law, $c^2=a^2+b^2-2abcos\theta$

We the plug in values: $(CB)^2+6^2+5^2-2*6*5*cos74$

and solve that CB is around 6.7

3 0

3 years ago

Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]$

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

3 0

3 years ago

Read 2 more answers