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tino4ka555 [31]
2 years ago
13

Find the product of

0-%204x%20%7B%7D%5E%7B3%7Dyz%20%7B%7D%5E%7B4%7D%29%20%20" id="TexFormula1" title="( - 12x ^{3}yz \:) and ( - 4x {}^{3}yz {}^{4}) " alt="( - 12x ^{3}yz \:) and ( - 4x {}^{3}yz {}^{4}) " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
Alex2 years ago
3 0

(-12x^3 yz)(-4x^3yz^4)\\\\=48x^6y^2 z^5

Rudik [331]2 years ago
3 0

Answer:

- 12{x}^{3} yz \times ( - 4 {x}^{3} y {z}^{4} )

48 {x}^{6}  {y}^{2}  {z}^{5}

Step-by-step explanation:

\frak{answered \: by....... \fbox{iuynsm}}

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BRAINLIESTT ASAP! PLEASE HELP ME :)
Lubov Fominskaja [6]

Answer: \sqrt{37}

=====================================

Work Shown:

6x-y = -3 is the same as 6x-y+3 = 0

6x-y+3 = 0 is the form Ax+By+C = 0 with A = 6, B = -1, C = 3.

The distance d that line is from the point (p,q) = (6, 2) is found by the following

d = \frac{A*p+B*q+C}{\sqrt{A^2+B^2}}\\\\d = \frac{6*6-1*2+3}{\sqrt{6^2+(-1)^2}}\\\\d = \frac{36-2+3}{\sqrt{36+1}}\\\\d = \frac{37}{\sqrt{37}}\\\\d = \frac{37\sqrt{37}}{37}\\\\d = \sqrt{37}\\\\d \approx 6.08276\\\\

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The answer is:
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