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anzhelika [568]
3 years ago
10

What is the expression in radical form? (3p^3q)^3/4

Mathematics
2 answers:
Makovka662 [10]3 years ago
8 0
\bf a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
(3p^3q)^{\frac{3}{4}}\implies \sqrt[4]{(3p^3q)^3}\implies \sqrt[4]{3^3p^{3\cdot 3}q^3}\implies \sqrt[4]{27p^9q^3}
\\\\\\
\sqrt[4]{27p^{4+4+1}q^3}\implies \sqrt[4]{27p^4p^4p^1q^3}\implies pp\sqrt[4]{27pq^3}\implies p^2\sqrt[4]{27pq^3}
disa [49]3 years ago
6 0
Look to the sky and close your eyes and wish 

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Alecsey [184]

Answer:

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4 0
3 years ago
How can i solve the equation 4(2x+7)
myrzilka [38]
Distribute so times what's outside by what's inside so 4×2x=8x and 4×7=28 then get x by itself by dividing on both sides 28÷8=3.5 your answer is x=3.5
3 0
3 years ago
Read 2 more answers
What is 56 divided by 3 1/2
Anton [14]

Answer:

Step-by-step explanation:

First you will need to change both of the numbers into improper fractions:

56/1 ÷ 7/2

Then do keep change flip. What that means is keep the first fraction the same, change the division symbol into a multiplication symbol, and flip the second fraction:

56/1 x 2/7

Then just multiply the two numerators by each other and the two denominators by each other:

56 x 2 = 112

7 x 1 = 7

112/7 would be the answer.

4 0
3 years ago
Can someone help me with 9 and 10? I’m really lost— thank you :)
puteri [66]
For 9, do 0.5*6*9.2*5 to get the total area of the visual triangles, add that to 0.5*6*4.1*5 to get the area of the pentagon and you should get 199.5 cm^2for 10, do 0.5*6.9*8*12 to get the total area of the figure and you should get 331.2 cm^2
8 0
4 years ago
Two landscapers must mow a rectangular lawn that measures 100 feet by 200 feet. Each wants to mow no more than half of the lawn.
Citrus2011 [14]

The total area of the complete lawn is (100-ft x 200-ft) = 20,000 ft².
One half of the lawn is  10,000 ft².  That's the limit that the first man
must be careful not to exceed, lest he blindly mow a couple of blades
more than his partner does, and become the laughing stock of the whole
company when the word gets around.  10,000 ft² ... no mas !

When you think about it ... massage it and roll it around in your
mind's eye, and then soon give up and make yourself a sketch ...
you realize that if he starts along the length of the field, then with
a 2-ft cut, the lengths of the strips he cuts will line up like this:

First lap:
       (200 - 0) = 200
       (100 - 2) = 98
       (200 - 2) = 198
       (100 - 4) = 96    

Second lap:
       (200 - 4) = 196
       (100 - 6) = 94
       (200 - 6) = 194
       (100 - 8) = 92   

Third lap:
       (200 - 8) = 192
       (100 - 10) = 90
       (200 - 10) = 190
       (100 - 12) = 88 

These are the lengths of each strip.  They're 2-ft wide, so the area
of each one is (2 x the length). 

I expected to be able to see a pattern developing, but my brain cells
are too fatigued and I don't see it.  So I'll just keep going for another
lap, then add up all the areas and see how close he is:

Fourth lap:
       (200 - 12) = 188
       (100 - 14) = 86
       (200 - 14) = 186
       (100 - 16) = 84 

So far, after four laps around the yard, the 16 lengths add up to
2,272-ft, for a total area of 4,544-ft².  If I kept this up, I'd need to do
at least four more laps ... probably more, because they're getting smaller
all the time, so each lap contributes less area than the last one did.

Hey ! Maybe that's the key to the approximate pattern !

Each lap around the yard mows a 2-ft strip along the length ... twice ...
and a 2-ft strip along the width ... twice.  (Approximately.)  So the area
that gets mowed around each lap is (2-ft) x (the perimeter of the rectangle),
(approximately), and then the NEXT lap is a rectangle with 4-ft less length
and 4-ft less width.

So now we have rectangles measuring

         (200 x 100),  (196 x 96),  (192 x 92),  (188 x 88),  (184 x 84) ... etc.

and the areas of their rectangular strips are
           1200-ft², 1168-ft², 1136-ft², 1104-ft², 1072-ft² ... etc.

==> I see that the areas are decreasing by 32-ft² each lap.
       So the next few laps are 
               1040-ft², 1008-ft², 976-ft², 944-ft², 912-ft² ... etc. 

How much area do we have now:

             After 9 laps,    Area =   9,648-ft²
             After 10 laps,  Area = 10,560-ft².

And there you are ... Somewhere during the 10th lap, he'll need to
stop and call the company surveyor, to come out, measure up, walk
in front of the mower, and put down a yellow chalk-line exactly where
the total becomes 10,000-ft².   


There must still be an easier way to do it.  For now, however, I'll leave it
there, and go with my answer of:  During the 10th lap.

5 0
3 years ago
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