| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.26 |
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Which of the following statements about math operations is incorrect?
all of these statements are correct |
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you can multiply monomials that have different variables and different exponents |
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you can subtract monomials that have the same variable and the same exponent |
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you can add monomials that have the same variable and the same exponent |
You can only add or subtract monomials that have the same variable and the same exponent. For example, 2a + 4a = 6a and 4a2 - a2 = 3a2 but 2a + 4b and 7a - 3b cannot be combined. However, you can multiply and divide monomials with unlike terms. For example, 2a x 6b = 12ab.
Solve 2b + 5b = 6b + 4x + 7 for b in terms of x.
| -6\(\frac{1}{2}\)x + 1 | |
| 6x + 7 | |
| 1\(\frac{1}{5}\)x + \(\frac{1}{5}\) | |
| \(\frac{1}{4}\)x - 1\(\frac{3}{4}\) |
To solve this equation, isolate the variable for which you are solving (b) on one side of the equation and put everything else on the other side.
2b + 5x = 6b + 4x + 7
2b = 6b + 4x + 7 - 5x
2b - 6b = 4x + 7 - 5x
-4b = -x + 7
b = \( \frac{-x + 7}{-4} \)
b = \( \frac{-x}{-4} \) + \( \frac{7}{-4} \)
b = \(\frac{1}{4}\)x - 1\(\frac{3}{4}\)
If the area of this square is 9, what is the length of one of the diagonals?
| 3\( \sqrt{2} \) | |
| 7\( \sqrt{2} \) | |
| 2\( \sqrt{2} \) | |
| 4\( \sqrt{2} \) |
To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:
a = s2
so the length of one side of the square is:
s = \( \sqrt{a} \) = \( \sqrt{9} \) = 3
The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:
c2 = a2 + b2
c2 = 32 + 32
c2 = 18
c = \( \sqrt{18} \) = \( \sqrt{9 x 2} \) = \( \sqrt{9} \) \( \sqrt{2} \)
c = 3\( \sqrt{2} \)
If BD = 10 and AD = 20, AB = ?
| 2 | |
| 7 | |
| 16 | |
| 10 |
The entire length of this line is represented by AD which is AB + BD:
AD = AB + BD
Solving for AB:AB = AD - BDBreaking apart a quadratic expression into a pair of binomials is called:
normalizing |
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squaring |
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deconstructing |
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factoring |
To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse.