Din class, to prove that the rsa decryption works, we used the fact that ed1 (mod p( prove that the decryption process still works if, instead of this, we assume ed 1 (mod lem(p-1,q- note: icni stands for least common multiple

u rich

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

p ÷ r=tan(x)

hence final answer is p ÷ r=tan(x).