A picture is x+20 cm in width and 2x-10 cm in length. The frame is a constant 5 cm wide from the edge of the picture to the outside of the
frame. What is the area of just the frame?

A = 1100cm^2

Step by step Explanation:

given the dimensions of width and length of the picture are x+20 by 2x-10 and the frame is a constant 5 cm wider from the edge of the picture to the frame, than the area of the frame is defined as (x+30)(2x)-(x+20)(2x-10) = 30x+200.

If the width is equal to the length which I assume is true if the width is constant.

than x+30=2x, which means x = 30.

if this is true than 30(30)+200 = 1100cm^2

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).