Prove the following using proof by contrapositive, proof by contradiction or an existence proof (a) if the gcd(a, b)1, then at most one of a or b is even (b) there are no positive real roots of 2 4x +5 (c) if r and s are both positive real numbers, then + st vf+ vs. (d) the only three consecutive odd natural numbers that are all prime are 3, 5, and (e) v5 is not a rational number (0) if r is a rectangle of area a, then there exists a square with the same area.

answer: 48

just remember to use pemdas:

parentheses

expressions

multiplication

division

addition

subtraction

answer: translate the circles so they share a common center point and dilate circle y by a scale factor of 3.

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .