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Andrei [34K]
2 years ago
9

Which expression is equivalent to RootIndex 3 StartRoot 64 a Superscript 6 Baseline b Superscript 7 Baseline c Superscript 9 Bas

eline EndRoot?
2 a b c squared (RootIndex 3 StartRoot 4 a squared b cubed c EndRoot)
4 a squared b squared c cubed (RootIndex 3 StartRoot b EndRoot)
8 a cubed b cubed c Superscript 4 Baseline (RootIndex 3 StartRoot b c EndRoot)
8 a squared b squared c cubed (RootIndex 3 StartRoot b EndRoot)
Mathematics
2 answers:
mart [117]2 years ago
4 0

Answer:

Which expression is equal to \sqrt[3]{64}a^6b^7c^9?

The correct answer is B.

                 4a^{2}b^{2}c^{3}(\sqrt[3]{b})

Step-by-step explanation:

Inside of the radical you have 64a^{6}. If you find the cube root of that, you get 4a^2. Go ahead and write that outside of the parenthesis:

                     4a^{2}\sqrt[x}\sqrt[3]({b^{7}c^{9}})

If you re-write what is inside of the radical, you get:

                4a^{2}(\sqrt[3]{b^{3}*b^{3}*b^{1}*c^{3}*c^{3}*c^{3}   }

Basically I expanded what was inside of the radical so I could find the cube roots of b^7 and c^9.

Now, take the cube root of b^7:

                       

                       4a^{2}b^{2} (\sqrt[3]b*c^{3}*c^{3}*c^{3}   })

Notice how I could only factor out the two "b^3" that were inside the radical symbol, and how I left the b^1 inside the radical symbol because I couldn't factor it out.

Let's now get the cube root of c^9. Since it's a perfect cube, there won't be any "c"s left inside of the radical symbol:

               

                            4a^{2}b^{2}c^{9}(\sqrt[3]b)

Makovka662 [10]2 years ago
3 0

Answer:

B

Step-by-step explanation:

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Billy had 55 stamps in his collection. billy’s older brother tommy had 73 stamps in his collection. how many stamps did the two
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Number of stamps Billy's older brother had = 73

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A recent poll of 124 randomly selected residents of a town with a population of 310 showed that 93 of them are opposed to a new
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1.) y=3
2.) y=13
     2y + y = 3y , 22+17= 39
     3y = 39 , divide 3 on each side (which cancels out the 3 with the y which leaves you with y)
3.) y=10
     3y - y = 2y , 35-15= 20
     2y=20 , do the same as # 2.
4.) y=4
     9y - 5y = 4y , 48-32= 16
     4y=16, do the same as #2 & #3
5.) (4,5)
     x + 3y = 19
   2x - 3y = -7
    solve for x first ; x + 2x= 3x , 19-7 = 12; 3x = 12 ; divide 3 ; x= 4. You know have to make the x's cancel out. so I will multiply -2 on the first equation. -2(x+3y=19)
     -2x - 6y = -38
      2x - 3y = -7
    Now solve for y ; -6y + -3y = -9y , -38 + -7 = -45; -9y=-45 ; divide by -9 (since they are both negative your answer will be positive) ; y= 5
6.) (6,8)
   x + 4y = 38
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solve for y first ; 4y - 3y = y , 38-30 = 8; y=8 (that simple!) Now you need to find a common multiple for 4 & 3 which is 12. So you will have to multiply each equation by either 4 or 3.
  (x + 4y = 38)*3      =     3x + 12y = 114         
 (-x - 3y = -30)*4     =     -4x - 12y = -120
solve for x ; 3x - 4x = -x , 114-120= -6 ; -x=-6. Since the x has a negative that means there's still a 1 there so -1x=-6 ; you will need to divide this which makes the 6 a positive; x=6


8 0
2 years ago
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