Maths

What is the form of the word analysis?

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What is the form of the word analysis?

A) plural possessive
B) plural
C) singular possessive
D) singular

Answer:

The form of the word analysis is singular. Thus, option (d) is correct.

A single research, report, assessment, or evaluation is referred to by the singular noun analysis. Analysis is written with a “-es” ending in the plural form rather than the singular “-is” ending.

A careful investigation of something difficult in order to comprehend it or identify its key characteristics is referred to as analysis. Because of its Greek etymology, the word analysis does not end with a -s or -es as is customary.

As a result, the significance of the word analysis are the aforementioned. Therefore, option (d) is correct.

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Convert 65 kg to lbs.

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Convert 65 kg to lbs.

Final Answer

To convert 65 kg to lbs, multiply by the conversion factor of approximately 2.20462. Thus, 65 kg is equal to about 143.30 lbs. This conversion is significant for understanding weight in different measurement systems.

Explanation

To convert kilograms (kg) to pounds (lbs), we can use the conversion factor where 1 kilogram is approximately equal to 2.20462 pounds.

Let’s go through the conversion step-by-step:

Identify the given weight: We have 65 kg.

Use the conversion factor: Multiply the weight in kilograms by the conversion factor. So, we calculate:

65 \mathrm{~kg} \times 2.20462 \frac{\mathrm{lbs}}{\mathrm{~kg}}=143.3003 \mathrm{lbs}

We can round this to approximately 143.30 lbs for simplicity.

Conclusion: Therefore, 65 kg is roughly equal to 143.30 lbs. This conversion is useful in various contexts, especially in areas where different measurement systems are utilized, like nutrition or fitness.

Examples & Evidence

For instance, if someone weighs 70 kg, we can convert that to pounds by calculating

70 kg×2.20462≈154.32 lbs

This shows how the conversion works with a different weight.

The conversion factor used is widely accepted in scientific and practical applications, affirming that 1 kg corresponds to about 2.20462 lbs.

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Where p and q are different positive primes.

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If (p + q)-1 (p-1 + q-1) = paqb, prove that a + b + 2 = 0, where p and q are different positive primes.

Solution:

It is given that

(p + q)-1 (p-1 + q-1) = paqb

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 77

By cross multiplication

p-1q-1 = paqb

By comparing the powers

a = – 1 and b = – 1

Here

LHS = a + b + 2

Substituting the values

= – 1 – 1 + 2

= 0

= RHS

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Find x and y, where a, b are different positive primes.

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If ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 74, find x and y, where a, b are different positive primes.

Solution:

It is given that

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 75

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 76

By comparing the base on both sides

2 = x

x = 2

– 4/3 = 2y

2y = – 4/3

By further calculation

y = – 4/3 × ½ = – 2/3

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If x4y2z3 = 49392, find the values of x, y and z

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If x4y2z3 = 49392, find the values of x, y and z, where x, y and z are different positive primes.

Solution:

It is given that

x4y2z3 = 49392

We can write it as

x4y2z3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 × 7

x4y2z3 = (2)4 (3)2 (7)3 ……. (1)

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 73

Now compare the powers of 4, 2 and 3 on both sides of equation (1)

x = 2, y = 3 and z = 7

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Evaluate x1/2. y-1. z2/3 when x = 9, y = 2 and z = 8.

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Evaluate x1/2. y-1. z2/3 when x = 9, y = 2 and z = 8.

Solution:

It is given that

x = 9, y = 2 and z = 8

We know that

x1/2. y-1. z2/3 = (9)1/2. (2)-1. (8)2/3

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 71

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 72

= 6

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If x = 103 × 0.0099, y = 10-2 × 110.

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If x = 103 × 0.0099, y = 10-2 × 110, find the value of ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 69

Solution:

It is given that

x = 103 × 0.0099, y = 10-2 × 110

We know that

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 70

= √9

= √(3 × 3)

= 3

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If a = 3 and b = – 2, find the values of:

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If a = 3 and b = – 2, find the values of:

(i) aa + bb

(ii) ab + ba.

Solution:

It is given that

a = 3 and b = – 2

(i) aa + bb = (3)3 + (-2)-2

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 67

(ii) ab + ba = (3)-2 + (-2)3

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 68

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Simplify and express with positive exponents:

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Simplify and express with positive exponents:

(3x2)0, (xy)-2, (-27a9)2/3.

Solution:

We know that

(3x2)0 = 1

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 66

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Prove that x = 2yz/y – z.

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If 2x = 3y = 12z, prove that x = 2yz/y – z.

Solution:

It is given that

2x = 3y = 12z

Consider

2x = 3y = 12z = k

Here

2x = k where 2 = (k)1/x

3y = k where 3 = (k)1/y

12z = k where 12 = (k)-1/z

We know that

12 = 2 × 2 × 3

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 64

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 65

Therefore, it is proved.

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