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Exponents and Powers

Exponents and Powers relates to CBSE/CBSE - Class 8/Maths

Top Tutors who teach Exponents and Powers

1
Deepak Joshi Class 8 Tuition trainer in Bangalore Featured
Ashok Nagar D' Souza Layout, Bangalore
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Indrajeet K. Class 8 Tuition trainer in Noida Featured
Sector 1 Greater Noida, Noida
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Saurav Mitra Class 8 Tuition trainer in Mumbai Featured
Nerul, Mumbai
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6 yrs of Exp
350per hour
Classes: Class 8 Tuition, Class 9 Tuition and more.

I have extensive experience teaching Mathematics to students of Classes 7 and 8, both in schools and coaching institutes. At this level, my focus...

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Megha A. Class 8 Tuition trainer in Lucknow Featured
LDA Colony, Lucknow
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8 yrs of Exp
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The experience of teaching Maths in grade 8 is wonderful.This is the time when students develop a phobia of Maths.Removing that phobia with my techniques...

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Jaikumar Veerwani Class 8 Tuition trainer in Surat Featured
Adajan Adajan Gam, Surat
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6 yrs of Exp
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Classes: Class 8 Tuition, Class 6 Tuition and more.

Certified in Professional and Life Skills from the All India Institute of Certified Professionals . With 6 years of experience teaching Class 8 students...

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Pooja Patel . Class 8 Tuition trainer in Indore Featured
Scheme No 114, Indore
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4 yrs of Exp
300per hour
Classes: Class 8 Tuition, Class 9 Tuition and more.

I’m Pooja Patel, an enthusiastic and result-oriented professional tutor at Wizdomi, with over 6 years of teaching experience in Mathematics and Science...

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Debabrato Chatterjee Class 8 Tuition trainer in Thane Featured
Thane West, Thane
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15 yrs of Exp
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This is Debabrato Chatterjee, an Maths and Science tutor with 13+ years of experience. I have taught students of IB, IGCSE, CBSE, ICSE, and Maharastra...

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Bindu C. Class 8 Tuition trainer in Ottapalam Featured
Ottapalam, Ottapalam
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15 yrs of Exp
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I am an experienced English teacher with over 20 years of teaching expertise, specializing in training students from Class 6 to Class 12. My passion...

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Bindu Davi Class 8 Tuition trainer in Enamavu Featured
Enamakkal, Enamavu
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15 yrs of Exp
350per hour
Classes: Class 8 Tuition, Class 6 Tuition and more.

I am a school teacher from Kerala , India. I have 15 years teaching experience in CBSE school. I have a degree in statistics. I am a certified trained...

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Ajay A S Class 8 Tuition trainer in Kota Featured
Talwandi Sector-B, Kota
Super Tutor
10 yrs of Exp
1250per hour
Classes: Class 8 Tuition, Engineering Entrance Coaching and more.

My teaching exp.11 year in most reputed coaching in india. Mentor of jee top rank air-12 yatis agrawal, 19, 28 many more in top 100 in jee adavnced....

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Exponents and Powers Questions

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Answered on 02/02/2024 Learn CBSE/CBSE - Class 8/Maths/Exponents and Powers

Pooja R. Jain

To simplify the expression (1/32)3(1/32)3, you raise the base to the power of the exponent: (1/32)3=13/323(1/32)3=13/323 13=113=1 323==32768 So, (1/32)3=1/32768(1/32)3=1/32768 Therefore, (1/32)3(1/32)3 simplifies to 1/327681/32768. read more

To simplify the expression (1/32)3(1/32)3, you raise the base to the power of the exponent:

(1/32)3=13/323(1/32)3=13/323

13=113=1

323==32768

So,

(1/32)3=1/32768(1/32)3=1/32768

Therefore, (1/32)3(1/32)3 simplifies to 1/327681/32768.

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Answered on 02/02/2024 Learn CBSE/CBSE - Class 8/Maths/Exponents and Powers

Pooja R. Jain

To evaluate the expression (5−1×82)/(2−3×10−1)(5−1×82)/(2−3×10−1), follow the order of operations (PEMDAS/BODMAS): Exponents and Roots Multiplication and Division (from left to right) Addition and Subtraction (from left to right) Let's... read more

To evaluate the expression (5−1×82)/(2−3×10−1)(5−1×82)/(2−3×10−1), follow the order of operations (PEMDAS/BODMAS):

  1. Exponents and Roots
  2. Multiplication and Division (from left to right)
  3. Addition and Subtraction (from left to right)

Let's calculate step by step:

5−1×82=5−1×64=5−64=−595−1×82=5−1×64=5−64=−59

2−3×10−1=2−3×0.1=2−0.3=1.72−3×10−1=2−3×0.1=2−0.3=1.7

Now, substitute these results back into the original expression:

−591.71.7−59

Finally, divide −59−59 by 1.71.7:

−591.7≈−34.705882351.7−59≈−34.70588235

So, the value of the given expression is approximately −34.71−34.71.

 
 
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Answered on 02/02/2024 Learn CBSE/CBSE - Class 8/Maths/Exponents and Powers

Pooja R. Jain

To find the value of mm in the equation 6m6−3=656−36m=65, follow these steps: Simplify the expression on the left side: 6m6−3=6m3=2m6−36m=36m=2m Now, set up the equation: 2m=652m=65 Solve for mm: m=652m=265 So, the value of mm for which 6m6−3=656−36m=65... read more

To find the value of mm in the equation 6m6−3=656−36m=65, follow these steps:

  1. Simplify the expression on the left side:

    6m6−3=6m3=2m6−36m=36m=2m

  2. Now, set up the equation:

    2m=652m=65

  3. Solve for mm:

    m=652m=265

So, the value of mm for which 6m6−3=656−36m=65 is m=652m=265 or m=32.5m=32.5.

 
 
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Answered on 02/02/2024 Learn CBSE/CBSE - Class 8/Maths/Exponents and Powers

Pooja R. Jain

Let's simplify the expression step by step: −1−1 First, perform the operations inside each set of parentheses: −1−1 −1−1 −1−1 Now, combine the terms inside the brackets: −1−1 To add the fractions, find a common denominator, which is 6: −1−1 −1−1 Finally,... read more

Let's simplify the expression step by step:

[(12−1)−(13−1)]−1[(21−1)(31−1)]−1

First, perform the operations inside each set of parentheses:

[(12−1)−(13−1)]−1[(21−1)(31−1)]−1

[(12−22)−(13−33)]−1[(2122)(3133)]−1

[(−12)−(−23)]−1[(2−1)(3−2)]−1

Now, combine the terms inside the brackets:

[−12+23]−1[2−1+32]−1

To add the fractions, find a common denominator, which is 6:

[−36+46]−1[6−3+64]−1

[16]−1[61]−1

Finally, take the reciprocal of 1661:

116=6611=6

Therefore, [(12−1)−(13−1)]−1=6[(21−1)(31−1)]−1=6

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Answered on 02/02/2024 Learn CBSE/CBSE - Class 8/Maths/Exponents and Powers

Pooja R. Jain

To simplify (−3)5×(53)5(−3)5×(35)5, let's break it down step by step: Evaluate (−3)5(−3)5: (−3)5=−3×−3×−3×−3×−3=−243(−3)5=−3×−3×−3×−3×−3=−243 Evaluate... read more

To simplify (−3)5×(53)5(−3)5×(35)5, let's break it down step by step:

  1. Evaluate (−3)5(−3)5:

    (−3)5=−3×−3×−3×−3×−3=−243(−3)5=−3×−3×−3×−3×−3=−243

  2. Evaluate (53)5(35)5:

    (53)5=5535==3555=

Now, multiply the results:

(−243)×(−243)×

The 243243 in the numerator and denominator cancels out:

(−243)×=−243×3125(−243)×=−243×3125

Now, calculate the product:

−243×3125=−−243×3125=

Therefore, (−3)5×(53)5(−3)5×(35)5 simplifies to −]−]

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