a+b+c+d+e = 30 (i) c+e = 14 (ii) d+b = 1 (iii) a = 2b-1 (iv) a+c = 10 (v) By (i)-(ii)-(iii): a = 15 then, by (iv): b = 8 and by (v): c = -5 Also, b = 8 so by (iii): d = -7 and then by (i), e = 19
the answer is a
the correct answer is the 15 number in the alphabet which is o
Assuming by "numbers" you mean "whole numbers":{10, 10, 12, 13, 15}If the five numbers are {a, b, c, d, e} with a ≤ b ≤ c ≤ d ≤ e, then:median = 12 → c = 12leaving only two numbers (a, b) ≤ 12.mode = 10 → two or more numbers equal 10 (which is less than 12) → a = b = 10The final two numbers (d, e) are not equal, both greater than 12, and such that the sum of all five numbers is 60.10 + 10 + 12 + d + e = 60 → d + e = 28 → d = 13, e = 15→ {a, b, c, d, e} = {10, 10, 12, 13, 15}[If "numbers" includes "decimal numbers" (ie numbers with a fractional part), then as long as d + e = 28 and 12 < d < e there are infinitely many solutions, eg {10, 10, 12, 12.5, 15.5}, {10, 10, 12, 13.75, 14.25}]
a(x5) + b(x5) + c(x5) + d(x5) + e(x5) = abcde(a+b+c+d+e) x5 = abcdeThis equation has at least 5 variables. To solve for all of them requires at least 4 more equations.
if a + b = e. e + d = e + d. that's the answer: e + d or it could be e + d = (a + b) + d but in simplest form, its e + d
a+b+c+d+e = 30 (i) c+e = 14 (ii) d+b = 1 (iii) a = 2b-1 (iv) a+c = 10 (v) By (i)-(ii)-(iii): a = 15 then, by (iv): b = 8 and by (v): c = -5 Also, b = 8 so by (iii): d = -7 and then by (i), e = 19
A B plus C plus D plus E plusorA B C D E plus plus plus plusor variations.
the answer is a
Let x = abc be a 3-digit number. The expanded for of x is 100*a + 10*b + c.And let y = de = 10*d + e be a 2-digit number. Then x*y = 1000*a*d + 100*a*e + 100*b*d + 10*b*e + 10*c*d + c*e = 1000*a*d + 100*(a*e + b*d) + 10*(b*e + c*d) + c*e.
d=17
E. D. Smith died on 1948-10-15.
the correct answer is the 15 number in the alphabet which is o
Transitive property of equality
10 events in a decathlon
Assuming by "numbers" you mean "whole numbers":{10, 10, 12, 13, 15}If the five numbers are {a, b, c, d, e} with a ≤ b ≤ c ≤ d ≤ e, then:median = 12 → c = 12leaving only two numbers (a, b) ≤ 12.mode = 10 → two or more numbers equal 10 (which is less than 12) → a = b = 10The final two numbers (d, e) are not equal, both greater than 12, and such that the sum of all five numbers is 60.10 + 10 + 12 + d + e = 60 → d + e = 28 → d = 13, e = 15→ {a, b, c, d, e} = {10, 10, 12, 13, 15}[If "numbers" includes "decimal numbers" (ie numbers with a fractional part), then as long as d + e = 28 and 12 < d < e there are infinitely many solutions, eg {10, 10, 12, 12.5, 15.5}, {10, 10, 12, 13.75, 14.25}]
10 Nodes A B C D E A-B B-C C-D D-E A-C A-D E-B E-C E-A B-D