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kolezko [41]
3 years ago
10

Which expression is equivalent to (x Superscript 27 Baseline y) Superscript one-third?

Mathematics
2 answers:
Kay [80]3 years ago
6 0

Answer:

um i cannot understand that ok ?

Step-by-step explanation:

sdas [7]3 years ago
6 0

Answer:

Answer is D

Step-by-step explanation:

On Edge 2020

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Fill in the blanks with two consecutive integers to complete the following inequality.
Eva8 [605]

Answer:

<h2>11 and 12</h2>

Step-by-step explanation:

You know that 134 is between 11^2 and 12^2 which is 121 and 144.

So the two integers would be 11 and 12

Hope this helps :)

3 0
3 years ago
What's the anwser to 5 6 7
ExtremeBDS [4]
5: n=135
6: n=3
7: n=214
3 0
3 years ago
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Plz answer these 2 i really need help I'm not all that smart
kati45 [8]
Answer: 141.7 cm3

Explanation: you multiple 6.1, 6.7, 10.4 then divide by 3 which equals 141.68. Round it and we’re left with 141.7 cm3.
8 0
3 years ago
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Can you please help me answer this question what's a+3
Inessa [10]

Answer:

a+3

Step-by-step explanation:

You cannot go any further in answering this question.  These two terms are not like terms so they cannot be combined.  Therefore, the answer is just a+3 itself.

7 0
3 years ago
Can someone help.
r-ruslan [8.4K]

Answer:

m∠C = 38°  (nearest whole number).

Step-by-step explanation:

To find the missing angle, use the Sine Rule.

<u>Sine Rule</u> (for angles)

\sf \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}

where:

  • A, B and C are the angles
  • a, b and c are the sides opposite the angles

Given:

  • A = 53°
  • a = 18
  • c = 14

To find m∠C, substitute the given values into the formula and solve for C:

\implies \sf \dfrac{\sin 53^{\circ}}{18}=\dfrac{\sin C}{14}

\implies \sf \sin C= \dfrac{14 \sin 53^{\circ}}{18}

\implies \sf C= \sin^{-1} \left( \dfrac{14 \sin 53^{\circ}}{18}\right)

\implies \sf C= 38.40096301...^{\circ}

Therefore, m∠C = 38°  (nearest whole number).

Learn more about the sine rule here:

brainly.com/question/27089170

brainly.com/question/27704834

5 0
2 years ago
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