If your body were a house, your skeleton would be the house's
a.
rooms
b.
roof
c.
wooden frame

Write a function called checkfactor that takes two arrays of positive numbers, firstnumberrow and secondnumberrow. checkfactor checks if the first entry in firstnumberrow is divisible by the first entry in secondnumberrow, and performs the same operation on the next array elements until all entries have been checked. all the numbers are positive and the number of entries in the arrays are the same. the function should return the identified divisible numbers in two row arrays named firstdivisible and seconddivisible. restrictions: branches or loops should not be used. the code must use the internal mod and logical functions. hint: the mod function should be used to determine if two numbers are divisible. ex: for num1 and num2 if mod(num1,num2) is 0, then the two numbers are divisible. this is matlab

Ac++an of of pi. in , pi is by : pi = 4 – 4/3 + 4/5 – 4/7 + 4/9 - 4/11 + 4/13 - 4/15 + 4/17 . ., to pi (9 ). , if 5 to pi, be as : pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 = 4 – 1. + 0.8 - 0. + 0. = 3.. atoofbe to pi?

The idea that, for each pair of devices v and w, there’s a strict dichotomy between being “in range” or “out of range” is a simplified abstraction. more accurately, there’s a power decay function f (·) that specifies, for a pair of devices at distance δ, the signal strength f(δ) that they’ll be able to achieve on their wireless connection. (we’ll assume that f (δ) decreases with increasing δ.) we might want to build this into our notion of back-up sets as follows: among the k devices in the back-up set of v, there should be at least one that can be reached with very high signal strength, at least one other that can be reached with moderately high signal strength, and so forth. more concretely, we have values p1 ≥ p2 ≥ . . ≥ pk, so that if the back-up set for v consists of devices at distances d1≤d2≤≤dk, thenweshouldhavef(dj)≥pj foreachj. give an algorithm that determines whether it is possible to choose a back-up set for each device subject to this more detailed condition, still requiring that no device should appear in the back-up set of more than b other devices. again, the algorithm should output the back-up sets themselves, provided they can be found.\