Can someone help me please

6/7 or -6/7

Step-by-step explanation:

Pythagoras.

c² = a² + b²

the trigonometric functions in a norm circle (radius = 1) express the side lengths of inscribed right-angled triangles.

sin(x) is one side (e.g. "a") "touching" the 90 degree angle.

cos(x) is the other side ("b") of the 90 degree angle.

and the radius (1) is the Hypotenuse (the side opposite of the 90 degree angle), which is called "c" in the formula of Pythagoras.

so, we have

1² = (sqrt(13)/7)² + b²

b² = 1² - (sqrt(13/7)² = 1 - 13/49 = 49/49 - 13/49 = 36/49

b = cos(x) = sqrt(36/49) = 6/7

as we don't know anything about the angle x except that it has a positive sin(x) value, the cos(x) value can be positive or negative depending on if x is an angle of the first or of the second quadrant in the norm circle.

in the first quadrant cos(x) is also positive, in the second quadrant cos(x) is negative.

the third and fourth quadrants do not apply, as sin(x) would be negative for both.

i'm sure goog can you get that answer

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.