State whether 7-the sq root of 2 is rational or irrational. explain your answer.

Irrational

I'm going to write 'x' in place of "square root of 2" just to save time. Let's consider the possibility that 7-x is rational. I'm going to show there's a contradiction which ultimately concludes that the expression is irrational.

If 7-x was rational, then 7-x = p/q for some integers p,q with q being nonzero. Solve for x to get

7-x = p/q

-x = (pq) - 7

x = -(p/q) + 7

x = 7 - (p/q)

x = (7q/q) - (p/q)

x = (7q-p)/q

So if 7-x were rational, then this forces x to be rational because 7q-p is an integer over q (also an integer). The format is rational = integer/integer. But this is where the contradiction lies. Remember that x is standing in place for "square root of 2", which is an irrational number. There is no way to write sqrt(2) as a ratio of two integers. So x cannot be both rational and irrational at the same time.

Therefore, 7-x must be irrational making 7-sqrt(2) to be irrational as well.

note: sqrt is shorthand for "square root"

another note: adding any rational number to an irrational number will always result in an irrational number.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

a: 1/6, j: 1/4

step-by-step explanation: you are welcome