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Law Incorporation [45]
3 years ago
8

Im just begin lazy ok

Mathematics
1 answer:
dalvyx [7]3 years ago
3 0

Answer:

I am  being lazy too

Step-by-step explanation:

81 x 1 = 81

81 x 2 = 162

81 x 3 = 243

81 x 4 = 324

81 x 5 = 405

81 x 6 = 486

81 x 7 = 567

81 x 8 = 648

81 x 9 = 729

81 x 10 = 810

81 x 11 = 891

81 x 12 = 972

81 x 13 = 1053

81 x 14 = 1134

81 x 15 = 1215

81 x 16 = 1296

81 x 17 = 1377

81 x 18 = 1458

81 x 19 = 1539

81 x 20 = 1620

81 x 21 = 1701

81 x 22 = 1782

81 x 23 = 1863

81 x 24 = 1944

81 x 25 = 2025

81 x 26 = 2106

81 x 27 = 2187

81 x 28 = 2268

81 x 29 = 2349

81 x 30 = 2430

81 x 31 = 2511

81 x 32 = 2592

81 x 33 = 2673

81 x 34 = 2754

81 x 35 = 2835

81 x 36 = 2916

81 x 37 = 2997

81 x 38 = 3078

81 x 39 = 3159

81 x 40 = 3240

81 x 41 = 3321

81 x 42 = 3402

81 x 43 = 3483

81 x 44 = 3564

81 x 45 = 3645

81 x 46 = 3726

81 x 47 = 3807

81 x 48 = 3888

81 x 49 = 3969

81 x 50 = 4050

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SOMEONW GIVE ME STEP BY STEP EQUATION I NEED TO SUBMIT THIS TONIGHT
8090 [49]

Answer:

no answer. Because 8-4 is not equal to 12

Step-by-step explanation:

8 0
3 years ago
A ball is thrown in the air from a ledge. Its height in feet is represented by
Strike441 [17]
D.6 is your answer
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4 0
3 years ago
Is 7/12 greater then 4/12
aev [14]
True.

7/12 is greater than 4/12.

Have a wonderful day! :)

6 0
3 years ago
Read 2 more answers
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
How do you write 18 tens in standard form
Liula [17]
1800 because the 18 cant fit in the tens
7 0
3 years ago
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