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What is half of three in fraction form?
Half of 3 means 3 divided by 2 or 3 multipled by 1/2. Thus 1/2 X 3 = 3/2.
What is the slope of the line that contains the points -1 2 and 3 3?
Points: (-1, 2) and (3, 3) Slope: 2-3/-1-3 = 1/4
How do you get the numbers one through fifty using only the digits 1234?
1=(4-3)÷(2-1)2=(4+2)-(3+1)3=3×2-4+14=4÷2+3-15=12-(3+4)6=(3×4)×1÷27=1+(3×4)÷28=4+2-1+39=41-3210=(1×4)+(3×2)11=42-3112=(4×3)×(2-1)13=12-3+414=(3×4)×1+215=(3×4)+1+216=2(3+4+1)17=3(4+2)-118=32-1419=14+3+220=4(2+3×1)21=(1+2)×(3+4)22=34-1223=4(3×2)-124=1×3×4×225=(4+1)×(3+2)26=2(4×3+1)27=1×3+2428=4(2×3+1)29=1+32-430=2(1+4)×331=34-1-232=2(1+3)×433=21+3×434=34(2-1)35=4+32-136=4(2+1)×337=14+2338=42-1-339=42-(3×1)40=43-1-241=43-(2×1)42=43-2+143=43(2-1)44=2+3(14)45=12×4-346=41+2+347=41+(3×2)48=24(3-1)49=I Don't Know50=13×4-2
What is the slope of the line through (-2 3) and (1 1)?
Points: (-2, 3) and (1, 1) Slope: -2/3
Find the two numbers whose product is 1 and whose sum is 1?
x + y = 1xy = 1y = 1 - xx(1 - x) = 1x - x^2 = 1-x^2 + x - 1 = 0 or multiplying all terms by -1;(-x^2)(-1) + (x)(-1) - (1)(-1) = 0x^2 - x + 1 = 0The roots are complex numbers. Use the quadratic formula and find them:a = 1, b = -1, and c = 1x = [-b + square root of (b^2 - (4)(a)(c)]/2a orx = [-b - square root of (b^2 - (4)(a)(c)]/2aSox = [-(-1) + square root of ((-1)^2 - (4)(1)(1)]/2(1)x = [1 + square root of (1 - 4]/2x = [1 + square root of (- 3)]/2 orx = [1 + square root of (-1 )(3)]/2; substitute (-1) = i^2;x = [1 + square root of (i^2 )(3)]/2x = [1 + (square root of 3)i]/2x = 1/2 + [i(square root of 3]/2 andx = 1/2 - [i(square root of 3)]/2Since we have two values for x, we will find also two values for yy = 1 - xy = 1 - [1/2 + (i(square root of 3))/2]y = 1 - 1/2 - [i(square root of 3)]/2y = 1/2 - [i(square root of 3)]/2 andy = 1 - [1/2 - (i(square root of 3))/2)]y = 1 - 1/2 + [i(square root of 3))/2]y = 1/2 + [i(square root of 3)]/2Thus, these numbers are:1. x = 1/2 + [i(square root of 3)]/2 and y = 1/2 - [i(square root of 3)]/22. x = 1/2 - [i(square root of 3)]/2 and y = 1/2 + [i(square root of 3)]/2Let's check this:x + y = 11/2 + [i(square root of 3)]/2 +1/2 - [i(square root of 3)]/2 = 1/2 + 1/2 = 1xy = 1[1/2 + [i(square root of 3)]/2] [1/2 - [i(square root of 3)]/2]= (1/2)(1/2) -(1/2)[i(square root of 3)]/2] + [i(square root of 3)]/2](1/2) - [i(square root of 3)]/2] [i(square root of 3)]/2]= 1/4 - [i(square root of 3)]/4 + [i(square root of 3)]/4 - (3i^2)/4; substitute ( i^2)=-1:= 1/4 - [(3)(-1)]/4= 1/4 + 3/4= 4/4=1In the same way we check and two other values of x and y.
