<span>Diameter of the sphere = 8 cm.
Therefore, the radius of the sphere = 8/2 = 4 cm
Volume of the sphere = 4/3 * 22/7 * 3 * 3 * 3 cu. cm.
The sphere is beaten and drawn into a wire, the wire is cylinder of the diameter = 0.2 cm
Area of the base of the cylinder = pi * r2 = 22/7 * 0.2 * 0.2 sq. cm
volume of the wire = pi * r2 * I = 22/7 * 0.2 * 0.2 * 1
Therefore, 22/7 * 0.2 * 0.2 * I = 4/3 * 22/7 * 4 * 4 * 4
Therefore I = 4/3 * 22/7 * 4 * 4 * 4 * 7/22 * 1/0.04
I = 256/3 * 1/0.04
I = 2113 cm</span>
Divide 3 by 5 and then multiply by 100 to get percentage.
3/5 = 0.60 x 100 = 60%
Answer: 60%
Answer:
Let AB be called x.
26/(18+x) = 14/x
14(18+x)=26x
252+14x=26x
252=12x
x=21
So, AB = 21
Let me know if this helps!
Answer:
-2x^4 +10x^3 -3x^2 +5x +27
Step-by-step explanation:
a) Using the distributive property, we can eliminate parentheses:
2x^4 + 7x^3 - 3x^2 + 7 - 4x^4 + 3x^3 + 5x + 20
And using the distributive property again, we can factor pairs of like terms:
= x^4(2 -4) +x^3(7 +3) -3x^2 +5x +(7 +20)
b)
= -2x^4 +10x^3 -3x^2 +5x +27
Answer:
650,000,000 student ID numbers are possible if the letters cannot be repeated.
Step-by-step explanation:
The order in which the digits or letters are placed is important, which means that the permutations formula is used to solve this question.
Permutations formula:
The number of possible permutations of x elements from a set of n elements is given by the following formula:
In this question:
2 letters from a set of 26(permutations, as letters cannot be repeated).
6 digits, each with 10 possible outcomes.
How many student ID numbers are possible if the letters cannot be repeated?
650,000,000 student ID numbers are possible if the letters cannot be repeated.
