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frutty [35]
3 years ago
10

Please Help! Cube Roots 3 square root 56 + 3 square root 189

Mathematics
1 answer:
leonid [27]3 years ago
8 0

Step-by-step explanation:

cubic root(56) + cubic root(189) =

= cubic root(7×8) + cubic root(27×7) =

= 2×cubic root(7) + 3×cubic root(7) = 5×cubic root(7) =

= 9.564655914...

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ollegr [7]
The answer would be 7 becuase if rounded
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3 years ago
-24-(-6)+(-31)<br> step by step<br> be sure to simplify first
trasher [3.6K]

Answer:

-49

Step-by-step explanation:

rule of thumb: whenever you have two negatives, they equal a positive! when you have one negative and one positive, it is a negative!

  1. let's start by going from left to right :))
  2. -24 -(-6) + (-31)
  3. -(-6) is actually + 6!
  4. -24 + 6 + (-31)
  5. +(-31) would be -31 because one - and one + is a -
  6. -24 + 6 - 31
  7. -18 - 31
  8. -49

comment or message me if you are still confused or have a question!

8 0
3 years ago
Amanda made $276 for 12 hours of work, At the same rate, how many hours would she have to work to make $115
Anit [1.1K]

Answer:

5 hours

Step-by-step explanation:

276/12= 23

115/23= 5

5 0
3 years ago
Read 2 more answers
Need brief explanation about why false is correct
anyanavicka [17]

<u>We are given the equation:</u>

(a + b)! = a! + b!

<u>Testing the given equation</u>

In order to test it, we will let: a = 2 and b = 3

So, we can rewrite the equation as:

(2+3)! = 2! + 3!

5! = 2! + 3!

<em>We know that (5! = 120) , (2! = 2) and (3! = 6):</em>

120 = 2 + 6

We can see that LHS ≠ RHS,

So, we can say that the given equation is incorrect

6 0
3 years ago
can you transform the square shown below into pieces that when put together form five small equal squares whose total area shall
Sunny_sXe [5.5K]

How about this (see attached image):

Use the four cuts as shown in the image (red lines).

Then assemble 5 equal squares by the numbers: 1 center square and the rest are pieced together using two pieces as shown. All five together add up to the same area as original square because we use all pieces.

The way one gets a hint toward a solution is to see how an area of a square of length 1 can be split into 5 equal square areas:

1^2 = 5\cdot x^2\\x = \frac{\sqrt{5}}{5} = \sqrt{\frac{2^2}{5^2}+\frac{1^2}{5^2}}

which indicates we need to find a a triangle with sides 2 and 1 to get the hypotenuse of the right length. That gave rise to the cut pattern (if you look carefully, there are triangles with those side lengths).

6 0
3 years ago
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