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baherus [9]
3 years ago
10

Hellpppp plz quickly

Mathematics
2 answers:
hammer [34]3 years ago
8 0
8 < \sqrt[3]{a} < 9
512 < a < 729

So, a can be any number between (512, 729).
∴ answer is (A) 679
OleMash [197]3 years ago
4 0
8 < ∛a < 9
We need to solve for a.
Let's get the 1/3th root of the whole equation:
\sqrt[1/3]{8} \ \textless \   \sqrt[1/3]{cubic-root-of-x} \ \textless \   \sqrt[1/3]{9}


\sqrt[1/3]{cubic-root-of-x} = x
\sqrt[1/3]{8} = 512
\sqrt[1/3]{9} = 712

So
\sqrt[1/3]{8} \ \textless \   \sqrt[1/3]{cubic-root-of-x} \ \textless \   \sqrt[1/3]{9}

512 < x < 729

Now from the values we got, 679 is between 512 and 729.

So  8 < ∛679 < 9 (∛679 ≈ 8.78) (Answer A)


Hope this helps! :D

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2√72 + 5√242 what is the answer for this question please can u tell me the right answer​
Radda [10]

Answer:

67\sqrt{2}

Step-by-step explanation:

Assuming you require to simplify the expression.

Using the rule of radicals

\sqrt{a} × \sqrt{b} ⇔ \sqrt{ab}

Simplifying the radicals

\sqrt{72} = \sqrt{36(2)} = \sqrt{36} × \sqrt{2} = 6\sqrt{2}

\sqrt{242} = \sqrt{121(2)} = \sqrt{121} × \sqrt{2} = 11\sqrt{2}

Then

2\sqrt{72} + 5\sqrt{242}

= 2(6\sqrt{2} ) + 5(11\sqrt{2} )

= 12\sqrt{2} + 55\sqrt{2}

= 67\sqrt{2}

6 0
3 years ago
Write 83 as a sum of tens and ones
Travka [436]
8 tens and 3 ones.


hope that helped
6 0
3 years ago
Read 2 more answers
Sum of two numbers is 84. If one of the numbers exceeds the other by 12, the numbers are
djverab [1.8K]

Answer:

54 and 30

Step-by-step explanation:

84÷2=42-12=30

42+12=54

5 0
3 years ago
In un rettangolo la somma delle sue dimensioni misura 154 cm e queste sono una i 5/6 dell'altra.Calcola l'area del rettangolo.
scZoUnD [109]
1043,83125      <span>Penso che sia la risposta</span>
6 0
3 years ago
What is the perimeter of quadrilateral DOBC?
3241004551 [841]

Answer:

Hey there if you meant the picture below, I wrote the answers for you

Step-by-step explanation:

Hope my you understand my handwriting

8+8+6+6

And that how to get your answer

By xBrainly

7 0
3 years ago
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