$\bf \cfrac{\sqrt[3]{144x}}{\sqrt[3]{y}}~~
\begin{cases}
144=2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\\
\qquad 2^3\cdot 18
\end{cases}\implies \cfrac{\sqrt[3]{2^3\cdot 18x}}{\sqrt[3]{y}}\implies \cfrac{2\sqrt[3]{ 18x}}{\sqrt[3]{y}}
\\\\\\
\cfrac{2\sqrt[3]{ 18x}}{\sqrt[3]{y}}\cdot \cfrac{\sqrt[3]{(18x)^2}}{\sqrt[3]{(18x)^2}}\implies \cfrac{2\sqrt[3]{(18x)(18x)^2}}{\sqrt[3]{(y)(18x)^2}}\implies \cfrac{2\sqrt[3]{(18x)^3}}{\sqrt[3]{18^2x^2y}}$

$\bf \cfrac{2(18x)}{\sqrt[3]{324x^2y}}~~
\begin{cases}
324=2\cdot 2\cdot 3\cdot 3\cdot 3\cdot 3\\
\qquad 12\cdot 3^3
\end{cases}\implies \cfrac{36x}{\sqrt[3]{12\cdot 3^3x^2y}}
\\\\\\
\cfrac{36x}{3\sqrt[3]{12x^2y}}\implies \cfrac{12x}{\sqrt[3]{12x^2y}}$

3 0

2 years ago

Read 2 more answers

NEED HELP Find values for a, b, c, and d so that the following matrix product equals the 2X2 identity matrix. Explain or show ho

ehidna [41]

We want to find the values of a, b, c, and d such that the given matrix product is equal to a 2x2 identity matrix. We will solve a system of equations to find:

b = 1
a = -1
c = -2
d = -1

<h3>Presenting the equation:</h3>

Basically, we want to solve:

$\left[\begin{array}{cc}-1&2\\a&1\end{array}\right]*\left[\begin{array}{cc}b&c\\1&d\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

The matrix product will be:

$\left[\begin{array}{cc}-b + 2&-c + 2d\\a*b + 1&a*c + d\end{array}\right]$

Then we must have:

-b + 2 = 1

This means that:

b = 2 - 1 = 1

b = 1

We also need to have:

a*b + 1 = 0

we know the value of b, so we just have:

a*1 + b = 0

a = -1

Now the two remaining equations are:

-c + 2d = 0

a*c + d = 1

Replacing the value of a we get:

-c + 2d = 0

-c + d = 1

Isolating c in the first equation we get:

c = 2d

Replacing that in the other equation we get:

-(2d) + d = 1

-d = 1

d = -1

Then:

c = 2d = 2*(-1) = -2

c = -2

So the values are:

b = 1
a = -1
c = -2
d = -1

If you want to learn more about systems of equations, you can read:

brainly.com/question/13729904

4 0

2 years ago

Charlemagne’s column covers the first 12 miles in 4 hours. He then increased the speed by 2 miles per hour. If the total distanc

Mashutka [201]

Answer:

12 miles in 4 hours is 3 mph

52-12=40 miles left

40 miles at 5 mph would take 8 hours

8+4=10

it took 10 hours to finish the trip.

8 0

2 years ago