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astraxan [27]
3 years ago
6

What is the value of 2 over 3 to the power of 0 to the power of -3

Mathematics
1 answer:
Schach [20]3 years ago
5 0

Answer:

((\frac{2}{3})^0)^{-3}=1

Step-by-step explanation:

We need to find the value of ((\frac{2}{3})^0)^{-3}

Solving:

We know, (a^b)^n = a^{b*n}

((\frac{2}{3})^{0*-3})

(\frac{2}{3})^0

a^0 = 1

so,

(\frac{2}{3})^0=1

So, the value of ((\frac{2}{3})^0)^{-3}=1

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Write 102,800,000 in scientific notation
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Answer:

1.028 × 10^8

Step-by-step explanation:

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Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997, However in march 1997 the price of the tea increased by 25% and
Irina18 [472]

Answer: tea = 15 rupees  per kg

sugar= 3 rupees per kg

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations with the information given:

<em>"Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997": </em>

2 t + 3 s =39 (a)

Where:

  • t= price of 1 kg of tea
  • s = price of 1 kg of sugar

<em>"in march 1997 the price of the tea increased by 25% (1.25)and the price of the sugar increased by 20%(1.20) and the same quantity of tea and sugar cost rupees 48.30. "</em>

2(t1.25)+3(s1.2) = 48.30 (b)

  • <em>Solving for t in (b) </em>

2t =39-3s

t = (39 -3s)/2

t = 19.5-1.5s

  • <em>Replacing the value of t in (b) </em>

2 x ((19.5-1.5s)1.25)+ 3 ( 1.2s) =48.30

2x ( 24.375 -1.875s) +3.6s =48.30

48.75 -3.75s+3.6s= 48.30

48.75-48.30 = 3.75s-3.6s

0.45= 0.15s

0.45/0.15 =s

3 =s

  • <em>Replacing the value of s in (a) </em>

2 t + 3 (3) =39

2 t + 9 =39

2 t =39 -9

2 t =30

t = 30/2

t= 15

Prices in january:

tea = 15 rupees per kg

sugar= 3 rupees per kg

Feel free to ask for more if needed or if you did not understand something.

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Answer:

rounded to the nearest ten thousand

2,034,627

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Which of these is the area of a sector of a circle with r = 18”, given that its arc length is 6π?
Igoryamba
22/7*18^2*6π/2π*18 = 169.65
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Add<br>(4-4i)+(3-2i)<br>Write as complex number in standard form​
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Answer: (4-4i)+(3-2i) = 7-6i

Step-by-step explanation:

To add or subtract two complex numbers, just add or subtract the corresponding real and imaginary parts. For instance, the sum of 4 -4i and 3 - 2i is 7 -6i. The numbers in standard form will be a + bi, where a is the real part and bi is the imaginary part.

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