Answer: 12
Step-by-step explanation:
Given the question :
Given the four digits 2, 4, 6, and 7, how many different positive two-digit integers can be formed using these digits if a digit may not be repeated in an integer?
The number of different positive two integer number can be obtained by:
P(4, 2) = 4P2
Recall:
nPr = n! / (n - r)!
4P2 = 4! / (4 - 2)!
4P2 = 4! / 2!
4P2 = (4 * 3 * 2 * 1) / ( 2 * 1)
4P2 = 24 / 2
4P2 = 12
Hence, 12 different positive two-digit integers can be formed using these digits if a digit may not be repeated in an integer
(i.) CA = πrl
CA = π (5*13)
CA = 65π
(ii.) TA = πrl + πr^2
TA = 65π + π (5^2)
TA = 65π + 25π
TA = 90π
(iii.) To get the height of the cone, you have to use the Pythagorean theorem. Plug in the radius for a and the slant height for c.
a^2 + b^2 = c^2
5^2 + b^2 = 13^2
25 + b^2 = 169 Height = 12
b^2 = 144
b = 12
(iv.) v = (1/3)πr^2h
v = (1/3)π(5^2)*12
v = (1/3)π(25*12)
v = (1/3)π*300
v = 100π
<span>189 / 6 = ( 180 + 6 ) / 6 = ( 180 / 6 ) + ( 6 / 6) = 30 + 1 = 31.</span>
91
Follow the order of operations.
3^4 + 2 * 5
81 + 2 * 5 (Exponent)
81 + 10 (Multiplication)
91 (Addition)
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Answer:
Two answers:
22 ≥ v
OR
v ≤ 22
Step-by-step explanation:
It's the same as solving for a variable in an equality. Imagine it as 17 = v - 5 for a moment. Simplify by subtracting 5 from both sides and re-add the inequality sign. This can be easily done since we are not multiplying or dividing, just adding and subtracting.
