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vazorg [7]
3 years ago
12

Use laws of exponents to simplify inside the parentheses of the given expression ((a ^ - 2 * b ^ 2)/(a ^ 2 * b ^ - 1)) ^ - 3

Mathematics
2 answers:
Vinvika [58]3 years ago
8 0

Answer:

its d

Step-by-step explanation:

ur welcome

Fudgin [204]3 years ago
4 0

Answer:

Step-by-step explanation:

[\frac{(a^{-2}*b^{2})}{(a^{2}*b^{-1})}]^{-3}=(a^{-2-2}*b^{2+1})^{-3}\\\\\\=(a^{-4}*b^{3})^{-3}=(a^{-4})^{-3}*(b^{3})^{-3}\\\\=a^{-4*-3}*b^{3*-3}=a^{12}*b^{-9}\\\\=\frac{a^{12}}{b^{9}}

Hint:  a^{-m}=\frac{1}{a^{m}}\\\\a^{m}*a^{n}=a^{m+n}\\\\(a^{m})^{n}=a^{mn}\\

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It costs $3.60 to buy one lb of candy. How much would it cost to purchase 5 lbs of candy?
Scilla [17]
33$ because 3.60•5lbs= 33$
3 0
3 years ago
Read 2 more answers
Q =3x2+5x-2 R = 2 – x2 S = 2x + 5 Find Q – [R + S] Show all of the steps.
matrenka [14]

Q=3x^2+5x-2\\R=2-x^2\\S=2x+5\\\\Q-[R+S]=(3x^2+5x-2)-[(2-x^2)+(2x+5)]\\\\=(3x^2+5x-2)-(2-x^2+2x+5)\\\\=(3x^2+5x-2)-[-x^2+2x+(2+5)]\\\\=(3x^2+5x-2)-(-x^2+2x+7)\\\\=3x^2+5x-2-(-x^2)-2x-7\\\\=3x^2+5x-2+x^2-2x-7\\\\=(3x^2+x^2)+(5x-2x)+(-2-7)\\\\=\boxed{4x^2+3x-9}

5 0
3 years ago
The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly
Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between \\ (\mu - 6\sigma) and \\ (\mu + 6\sigma). It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027

For those with defects <em>operating at a 6 sigma level, </em>the probability is:

\\ 1 - p = 1 - 0.999999998027 = 0.000000001973

Similarly, for finding <em>no defects</em> in a 5 sigma level, we have:

\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697.

The probability of defects is:

\\ 1 - p = 1 - 0.999999426697 = 0.000000573303

Well, the defects present in a six sigma level and a five sigma level are, respectively:

\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}

\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}  

Then, comparing both fractions, we can confirm that a <em>6 sigma level is markedly different when it comes to the number of defects present:</em>

\\ {6\sigma} \approx \frac{2}{10^9} [1]

\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9} [2]

Comparing [1] and [2], a six sigma process has <em>2 defects per billion</em> opportunities, whereas a five sigma process has <em>600 defects per billion</em> opportunities.

8 0
3 years ago
Solve for " x " <br><br><img src="https://tex.z-dn.net/?f=x%20%3D%20%2042%20-%209" id="TexFormula1" title="x = 42 - 9" alt="x =
disa [49]

\implies {\blue {\boxed {\boxed {\purple {\sf {x\:=\:33}}}}}}

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}

x = 42 - 9

➼\:x = 33

Therefore, the value of x is 33.

\sf\pink{Thank\:you. \:ヅ}

7 0
3 years ago
Read 2 more answers
It takes
kati45 [8]

Answer:

I don't really know if this is a trick question or not but, is 6363 minutes the answer?

Step-by-step explanation:

8 0
3 years ago
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