Answers:Part A: The value of x is 0.Part B: X can be any real number.
In Part A, you have to first evaluate 7^2. This is 49. Now, write the equation 49^x = 1. We know that if you raise any number to 0, then the answer is 1.
In Part B, you have to first evaluate 7^0, that is 1. Now, we have the equation 1^x = 1. In this case, 1 raised to any exponent is still only 1. Imagine 1^17, this would be 1 times itself 17 times or just 1.
Therefore any number will work in Part B.
To solve this problem you must use Permutation:
You want to made 4 digit numbers with the digits 1, 2, 3, 4, 5, 6 and<span> no digit can be used more than once. So, we have:
(A,B,C,D)
A can variate in six (6) differents forms.
B can variate in (6-1)=5 differents forms.
C can variate 4 digits.
D can variate 3 digits.
Keeping this on mind, we have:
=6!/2!
=6*5*4*3
=360
</span> The answer is: 360
<span><span><span>1. a = b means a is equal to b.
2. a ≠ b means a does not equal b.Operations1. Addition: If a = b then a + c = b + c.</span></span><span><span>2. Subtraction: If <span>a = </span>b then a – c = b– c.
3. Multiplication: If a = b then ac = bc.
<span>4. Division: If a = b and <span>c ≠ </span>0 then a/c = b/c.</span></span></span></span>
X = 2y + 8
x - y = 25
2y + 8 - y = 25
y = 25 - 8
y = 17
x = 2y + 8
x = 2(17) + 8
x = 34 + 8
x = 42
the larger number (x) = 42...and the smaller number (y) = 17
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