In terms of a e the distance formula to calculate the length of PQ in terms
Prove that PQ = /2 - 2(cosacosB + sinasin)​

step-by-step explanation:

compare

it would open into a square depending on the type of envelope i’m not sure about the second part

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

ending amount = beginning amount / (2^n)

where "n" = number of half-lives (or elapsed time / half-life)

ending amount = 50 mg / (2 ^10 / 5.3)

ending amount = 50 mg / (2 ^ 1.8867924528)

ending amount = 50 m / 3.6981210597

ending amount = 13.5203794557 mg

or about 13.5 mg remain

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