Certainly! Let's address each problem step-by-step:
48. How many consecutive odd numbers will be needed to obtain the sum as [tex]\(10^3\)[/tex]?
To determine how many consecutive odd numbers we need to sum to obtain 1000, we start by understanding that the sum of the first [tex]\(n\)[/tex] odd numbers is given by [tex]\(n^2\)[/tex]. Therefore, to find the number of odd numbers, [tex]\(n\)[/tex], such that their sum is close to 1000, we test increasing values of [tex]\(n\)[/tex]:
[tex]\[ 1 + 3 + 5 + \ldots + (2n-1) = n^2 \][/tex]
If we look for the closest square [tex]\( n^2 \)[/tex] to 1000:
For [tex]\( n = 31 \)[/tex], [tex]\( 31^2 = 961 \)[/tex] (which is less than 1000),
For [tex]\( n = 32 \)[/tex], [tex]\( 32^2 = 1024 \)[/tex] (which is slightly more than 1000).
Therefore, to get a sum close to [tex]\( 10^3 \)[/tex], we need 32 consecutive odd numbers. The correct answer is:
[tex]\[ (4) 32 \][/tex]
49. The value of [tex]\( (68.921)^{2/3} \)[/tex] is:
We need to calculate the value of [tex]\( 68.921^{2/3} \)[/tex]. Raising a number to the power of [tex]\( \frac{2}{3} \)[/tex] involves finding the cube root of the number and then squaring the result:
[tex]\[
68.921^{2/3} \approx 18.41
\][/tex]
Thus, the correct answer is:
[tex]\[ (2) 18.41 \][/tex]
50. What is the least number by which 12096 is divided, to get the quotient as a perfect cube number?
To find the divisor which makes the quotient a perfect cube, we test different numbers. By checking:
[tex]\(
12096 \div 6 = 2016 \implies 2016^{1/3} \approx 12.6 \quad (\text{not an integer})
\\
12096 \div 7 = 1728 \implies 1728^{1/3} = 12 \quad (\text{perfect cube})
\\
12096 \div 5 = 2419.2 \implies 2419.2^{1/3} \approx 13.4 \quad (\text{not an integer})
\\
12096 \div 8 = 1512 \implies 1512^{1/3} \approx 11.37 \quad (\text{not an integer})
\)[/tex]
The least number that gives a perfect cube quotient is 7, with the quotient being 1728. Therefore, the correct answer is:
[tex]\[ (2) 7 \][/tex]
51. Cube root of the quotient of two positive integers is:
The cube root of any positive number remains positive. Therefore, the cube root of a quotient of two positive integers is also positive.
Therefore, the correct answer is:
[tex]\[ (1) Positive \][/tex]
