166
I guess you just subtract the numbers
297-131=
Or there is something am I missing on here
Answer:
Step-by-step explanation:
This question is asking us to find where sin(2x + 30) has a sin of 1. If you look at the unit circle, 90 degrees has a sin of 1. Mathematically, it will be solved like this (begin by taking the inverse sin of both sides):
![sin^{-1}[sin(2x+30)]=sin^{-1}(1)](https://tex.z-dn.net/?f=sin%5E%7B-1%7D%5Bsin%282x%2B30%29%5D%3Dsin%5E%7B-1%7D%281%29)
On the left, the inverse sin "undoes" or cancels the sin, leaving us with
2x + 30 = sin⁻¹(1)
The right side is asking us what angle has a sin of 1, which is 90. Sub that into the right side:
2x + 30 = 90 and
2x = 60 so
x = 30
You're welcome!
Answer:
0.13 tax
Step-by-step explanation:
0.044*3=0.132
Answer and Step-by-step explanation:
(a) Given that x and y is even, we want to prove that xy is also even.
For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.
(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:
Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.
(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.
