20y=-5x+10
y=-1/4x+1/2
y-3=-1/4(x-8)
y-3=-1/4x+2
d. y = -1/4x + 5
Number of choices per question = 5
Number of wrong answers = 4
P(wrong answer) = 4/5
Answer: 4/5
Answer:
c. 3a^2b^11/2
Step-by-step explanation:
The applicable rule of exponents is ...
(a^b)/(a^c) = a^(b-c)
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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.
The number of passwords that would be allowed if users were allowed to reuse the letters and numbers is 30891577600
<h3>How to determine the number of passwords</h3>
The given parameters are:
Password length = 8
Letters = 6
Numbers = 2
There are 26 letters and 10 digits.
Since the letters and the numbers can be repeated, then the number of passwords is:
Count = 26^6 * 10^2
Evaluate
Count = 30891577600
Hence, the number of passwords that would be allowed if users were allowed to reuse the letters and numbers is 30891577600
Read more about combination at:
brainly.com/question/11732255
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