Solve for a in the figure shown.

Mathematics

2 answers:

aleksandrvk [35]2 years ago

7 0

Answer: 58 is the answer

hope this help and here my steps use a mark or a pin

and line it with the line if it's wrong sorry for the wrong

answer

Molodets [167]2 years ago

6 0

Check the picture below.

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Use strong mathematical induction to prove the existence part of the unique factorization of integers theorem (Theorem 4.4.5). I

valentina_108 [34]

Answer:

Lets say that P(n) is true if n is a prime or a product of prime numbers. We want to show that P(n) is true for all n > 1.

The base case is n=2. P(2) is true because 2 is prime.

Now lets use the inductive hypothesis. Lets take a number n > 2, and we will assume that P(k) is true for any integer k such that 1 < k < n. We want to show that P(n) is true. We may assume that n is not prime, otherwise, P(n) would be trivially true. Since n is not prime, there exist positive integers a,b greater than 1 such that a*b = n. Note that 1 < a < n and 1 < b < n, thus P(a) and P(b) are true. Therefore there exists primes p1, ...., pj and pj+1, ..., pl such that

p1*p2*...*pj = a

pj+1*pj+2*...*pl = b

As a result

n = a*b = (p1*......*pj)*(pj+1*....*pl) = p1*....*pj*....pl

Since we could write n as a product of primes, then P(n) is also true. For strong induction, we conclude than P(n) is true for all integers greater than 1.

5 0

2 years ago

What is the area of an isosceles trapezoid, if the lengths of its bases are 16 cm and 30 cm and the diagonals are perpendicular

kicyunya [14]

Call the point of intersection of the diagonals point X.

Each base is the hypotenuse of an isosceles right triangle whose sides are the diagonals and whose 90° angle is at X. The altitude of that triangle (⊥ distance to the base from X) is half the length of the hypotenuse. Then the height of the trapezoid is half the sum of the base lengths.

The area of the trapezoid is the product of the height and half the sum of the base lengths, hence is the square of half the sum of the base lengths.

... Area = ((16 cm +30 cm)/2)² = (23 cm)² = 529 cm²