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3 years ago

What is the correct answer to this multiple choice question? Please help!

Mathematics

1 answer:

erik [133]3 years ago

4 0

Answer:

$\sqrt[3]{4}$

Step-by-step explanation:

To solve this we need to take the square root of both sides with a certain degree.

$4^6 = a^{18}$

We can take the root of both sides with a degree of 18.

$\sqrt[18]{4^6} = \sqrt[18]{a^{18}} \\(4^6)^{\frac{1}{18} } = a\\a = 4^{\frac{6}{18} } = 4^{\frac{1}{3} } = \sqrt[3]{4}$

It's helpful to understand that when a number has its square root taken with some degree n. The square root can be represented as just the value raised to the (1/n) power.

It's also helpful to understand that when you raise something to the power of another power, you can simply multiply the powers together. For instance (2^3)^5 = 2^15

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Jacob solves the system of equations by forming a matrix equation.

Lyrx [107]

Simultaneous equations can be solved using inverse matrix operation.

The complete steps of Jacob's solution are:

$\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$

We have:

$4x + y = 2$

$-2x + 4y = -22$

Calculate the determinant of $\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

$|A| = 4 \times 3 -1 \times -2$

$|A| = 12 +2$

$|A| = 14$

So, the inverse matrix becomes

$A = \frac{1}{14}\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

Replace the first column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of x

$x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right]$

So, we have:

$x = \frac{1}{14}(2 \times 3 - 1 \times -22)$

$x = \frac{1}{14}(6 +22)$

$x = \frac{1}{14}(28)$

$x = 2$

Replace the second column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of y

$y = \frac{1}{14}\left[\begin{array}{cc}4&2\\-2&-22\end{array}\right]$

So, we have:

$y = \frac{1}{14}(4 \times -22 - 2 \times -2)$

$y = \frac{1}{14}(-88 +4)$

$y = \frac{1}{14}(-84)$

$y = -6$

Hence, the complete process is:

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$

Read more about matrices at:

brainly.com/question/11367104

6 0

3 years ago

Terri is reading a book that is 108 pages long. The book contains a 12-page introduction and 6 chapters. Each chapter has the sa

andrew-mc [135]

Answer:

Step-by-step explanation:

subtract then divide

3 0

3 years ago

Read 2 more answers

Write the equation For a Line that intersects points(1,3) And (2,5) which one is a.y=x+2 B.y=2x+1. C.y=3x-3. D.Y=-x+3 For me is

vlada-n [284]

$(1,3) ,\ \ (2,5) \\ \\ \\ First \ find \ the \ slope \ of \ the \ line \ thru \ the \ points \: \\ \\ m= \frac{y_{2}-y_{1}}{x_{2}-x_{1} } \\ \\m=\frac{5-3}{2-1} = \frac{2}{1}=2 \\ \\Now \ use \ y = mx + b \ with \ either \ point \ to \ find \ b, \ the \ y-intercept \ : \\ \\ y=mx+b \\ \\3=2 \cdot 1+b\\ \\3= 2 +b \\ \\b=3-2=1 \\ \\ y=2x+1 \ is \ the \ answer \ B$