**Given information:**
Vector A is given as A = 2i + 4j - 5k.

**Finding the magnitude of the vector:**
The magnitude of a vector can be found using the formula:
|A| = sqrt(Ax^2 + Ay^2 + Az^2)

Here, Ax, Ay, and Az represent the components of the vector A in the x, y, and z directions, respectively.
Plugging in the values from the given vector A, we have:
|A| = sqrt((2^2) + (4^2) + (-5^2))
= sqrt(4 + 16 + 25)
= sqrt(45)
= 6.71 (approx.)

Therefore, the magnitude of vector A is approximately 6.71.

**Finding the direction cosines of the vector:**
The direction cosines of a vector can be found using the formula:
cosθx = Ax / |A|
cosθy = Ay / |A|
cosθz = Az / |A|

Here, θx, θy, and θz represent the angles that the vector A makes with the x, y, and z axes, respectively.
Plugging in the values from the given vector A and the magnitude of A, we have:
cosθx = 2 / 6.71
= 0.298 (approx.)
cosθy = 4 / 6.71
= 0.597 (approx.)
cosθz = -5 / 6.71
= -0.746 (approx.)

Therefore, the direction cosines of vector A are approximately 0.298 in the x-direction, 0.597 in the y-direction, and -0.746 in the z-direction.

**Finding the angle the vector makes with respect to the axes:**
The angle that the vector A makes with the x, y, and z axes can be found using the formula:
θx = cos^(-1)(cosθx)
θy = cos^(-1)(cosθy)
θz = cos^(-1)(cosθz)

Using the direction cosines obtained earlier, we have:
θx = cos^(-1)(0.298)
= 71.67° (approx.)
θy = cos^(-1)(0.597)
= 54.16° (approx.)
θz = cos^(-1)(-0.746)
= 135.36° (approx.)

Therefore, the angle that vector A makes with the x-axis is approximately 71.67°, with the y-axis is approximately 54.16°, and with the z-axis is approximately 135.36°.

A= 2i+4j-5k then magnitude=√2^2 +4^2 +5^
= √25+25
=5√2
=7 unit