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3 years ago

Simplify (9^2/3*9^4/2)/9^1/2

Mathematics

1 answer:

JulsSmile [24]3 years ago

6 0

Answer:

$\frac{9^{5}}{3}$

Step-by-step explanation:

$\frac{9^{2}}{3} * \frac{9^{4}}{2} = \frac{9^{6}}{6}\\\\\frac{\frac{9^{6}}{6}}{\frac{9^{1}}{2}} \\\\\frac{2*9^{6}}{6*9^{1}}\\\\\frac{9^{5}}{3}$

$\frac{9^{5}}{3} = 19,683$

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A bag contains 4 white beads, 6 red beads, and 5 yellow beads. One bead is selected and not replaced. Then another bead is selec

lara31 [8.8K]

Answer:

$\frac{24}{210}$

Step-by-step explanation:

Add them all up 15 if one is taken out and no replaced OR put back in, then the second draw will have 14 beads.

4 x 6 = 24

15 x 14 = 210

7 0

3 years ago

Roger gets $40 per day as wages and $4.50 as commission for every pair of shoes he sells in a day. His daily earnings goal is $1

Mila [183]

Answer:

40+4.50x+112

16 shoes daily to make his goal

Step-by-step explanation:

to find how much shoes he would have to sell take away 40 from both sides of the equation which would leave you with 4.50x=72 next divide 4.50 out of both sides of the equation x=16 so he would have to sell 16 pairs of shoes a day to meet his goal

3 0

3 years ago

Read 2 more answers

Hfuuf help me plsssss

denpristay [2]

Answer:

a) 26

b) 26/3

Step-by-step explanation:

1) First, you have to turn 4 1/3 into an improper fraction, so you get 13/3

2) Then you do 13/3 *6/1 =78/3 (so you multiply both numerators and denominators)

3) Lastly, 78/3 can be simplified as 26

1) First you turn both fractions into improper fractions, so you get 13/5 and 10/3

2) Then you do 13/5*10/3 (so you multiply both numerators and denominators)

3) You get 130/15, which can be simplified as 26/3

4 0

3 years ago

The surface area of the cube is 54 square feet. What is the length, in feet, of one edge of the cube?

Brut [27]

Answer:

3 ft

Step-by-step explanation:

A cube is made of 6 equal faces each of which is a square with the same side length. The surface area of the cube is the surface area of one side multiplied by 6. We know the surface area is 54 so divide this by 6.

54/6 = 9 ft^2

Since the surface must be a square, take the square root of 9 which is 3 ft. The length of one edge of the cube is 3 ft.

5 0

3 years ago

Please help!!<br> Write a matrix representing the system of equations

frozen [14]

Answer:

$(4, -1, 3)$

Step-by-step explanation:

We have the system of equations:

$\left\{ \begin{array}{ll} x+2y+z =5 \\ 2x-y+2z=15\\3x+y-z=8 \end{array} \right.$

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

$\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}$

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

$\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}$

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

$\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}$

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

$\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}$

Now, we can multiply R₂ by -1/5. This yields:

$\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}$

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}$

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}$

We can now reduce R₃ by multiply it by -1/4:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

$\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

And finally, we can eliminate the second 1 by adding -(R₃):

$\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

Therefore, our solution set is $(4, -1, 3)$

And we're done!

3 0

3 years ago