Answer:
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I
1. 3×3×3×3 = 81
2. 2²a³
3. (-5)³ (-5×-5×-5) = -125
4. 1
5. a^m-n
II
1. 1.6807× 10⁴
2. (7×7×7) × (2×2×2×2×2×2) = 21952
3. 32²
4. 2187
5. 803025
<h3>
Answer: 14.85 + c = 20</h3>
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Reason:
To compute the change, we subtract the amount William gives the cashier (the $20) and the price of the book ($14.85)
The change c is
c = 20-14.85
To rearrange this equation, we can add 14.85 to both sides to end up with 14.85 + c = 20
In other words, solving that equation in bold leads to c = 20-14.85 = 5.15 which is the change William is given.
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Extra info (optional section)
If the cashier wanted to mentally figure out the change in his/her head, then notice the jump from 85 cents to 100 cents is 15 cents. That means we go from 14.85 to 15.00
Then the jump from $15 to $20 is an extra 5 dollars. Overall, the total change given back to William is 0.15+5 = 5.15 dollars.
The correct answer would be A) X^6 Y^5
Answer: 3/2
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Explanation:
Pick any two points on the red line. I'll pick (0,1) and (2,4)
Those coordinates will be plugged into the slope formula below
m = (y2 - y1)/(x2 - x1)
m = (4 - 1)/(2 - 0)
m = 3/2
This means we go up 3 and over to the right 2 when traveling along this red line.
Y = |x² - 3x + 1|
y = x - 1
|x² - 3x + 1| = x - 1
|x² - 3x + 1| = ±1(x - 1)
|x² - 3x + 1| = 1(x - 1) or |x² - 3x + 1| = -1(x - 1)
|x² - 3x + 1| = 1(x) - 1(1) or |x² - 3x + 1| = -1(x) + 1(1)
|x² - 3x + 1| = x - 1 or |x² - 3x + 1| = -x + 1
x² - 3x + 1 = x - 1 or x² - 3x + 1 = -x + 1
- x - x + x + x
x² - 4x + 1 = -1 or x² - 2x + 1 = 1
+ 1 + 1 - 1 - 1
x² - 4x + 1 = 0 or x² - 2x + 0 = 0
x = -(-4) ± √((-4)² - 4(1)(1)) or x = -(-2) ± √((-2)² - 4(1)(0))
2(1) 2(1)
x = 4 ± √(16 - 4) or x = 2 ± √(4 - 0)
2 2
x = 4 ± √(12) or x = 2 ± √(4)
2 2
x = 4 ± 2√(3) or x = 2 ± 2
2 2
x = 2 ± √(3) or x = 1 ± 1
x = 2 + √(3) or x = 2 - √(3) or x = 1 + 1 or x = 1 - 1
x = 2 or x = 0
y = x - 1 or y = x - 1 or y = x - 1 or y = x - 1
y = (2 + √(3)) - 1 or y = (2 - √(3)) - 1 or y = 2 - 1 or y = 0 - 1
y = 2 - 1 + √(3) or y = 2 - 1 - √(3) or y = 1 or y = -1
y = 1 + √(3) or y = 1 - √(3) (x, y) = (2, 1) or (x, y) = (0, -1)
(x, y) = (2 ± √(3), 1 ± √(3))
The solution (0, -1) can be made by one function (y = x - 1) while the solution (2 ± √(3), 1 ± √(3)) can be made by another function (y = |x² - 3x + 1|). So the solution (2, 1) can be made by both functions, making the two solutions equal.
