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First of all I woulod switch it around so it would be 36-4+k. Since the negative sign is the same as subtracting. So the final answers going to be 32+k.

Q: What is the answer for -4 plus 36 plus k?

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2k + 5

The value of X in the equation 2x plus 36 = 4 is -16.

1 + 4 + 16 + 25 + 36 + 49 = 131

36 divided by 4 is 9 and 9 plus 22 equels 31

If: x^2+y^2 = k and y = 3x+1 Then: x^2+(3x+1)^2 = k => x^2+9x^2+6x+1 = k Collecting like terms and subtracting k from both sides: 10x^2+6x+(1-k) = 0 For the line to touch the curve at one point the discriminant of b^2-4(ac) must = 0 So: 36-4*10*(1-k) = 0 => 36-40+40k = 0 => 40k = 40-36 => 40k = 4 Therefore: k = 4/40 and in its lowest terms 1/10

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If: y = kx+1 and y = 3x2-4x+4 Then: 3x2-4x+4 = kx+1 So: 3x2-4x-kx+3 = 0 For the line to be tangent to the curve the discriminant of b2-4ac must = 0 So when: -4*3*3 = -36 then (-4-k)2 must = 36 So it follows: (-4-k)(-4-k) = 36 => k2+8k-20 = 0 Solving the quadratic equation: k = 2 or k = -10

is multiplication , so you multiply like k(9)=36 then k would equal k=4 because you divide 9 by 36 and it equals =4

2k + 5

The value of X in the equation 2x plus 36 = 4 is -16.

1 + 4 + 16 + 25 + 36 + 49 = 131

36 divided by 4 is 9 and 9 plus 22 equels 31

sqrt(36) + 4 = 6 + 4 = 10

If: x^2+y^2 = k and y = 3x+1 Then: x^2+(3x+1)^2 = k => x^2+9x^2+6x+1 = k Collecting like terms and subtracting k from both sides: 10x^2+6x+(1-k) = 0 For the line to touch the curve at one point the discriminant of b^2-4(ac) must = 0 So: 36-4*10*(1-k) = 0 => 36-40+40k = 0 => 40k = 40-36 => 40k = 4 Therefore: k = 4/40 and in its lowest terms 1/10

36

32 + 4 = 36

They are members of the infinite set of numbers of the form 36*k where k is an integer.

If: -8x+4 = 36 then the value of x works out as -4 Check: -8*(-4)+4 = 36