| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.18 |
| Score | 0% | 64% |
Which of the following expressions contains exactly two terms?
polynomial |
|
quadratic |
|
binomial |
|
monomial |
A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms.
Which of the following statements about math operations is incorrect?
you can subtract monomials that have the same variable and the same exponent |
|
you can add monomials that have the same variable and the same exponent |
|
all of these statements are correct |
|
you can multiply monomials that have different variables and different exponents |
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.
Which of the following is not required to define the slope-intercept equation for a line?
slope |
|
y-intercept |
|
\({\Delta y \over \Delta x}\) |
|
x-intercept |
A line on the coordinate grid can be defined by a slope-intercept equation: y = mx + b. For a given value of x, the value of y can be determined given the slope (m) and y-intercept (b) of the line. The slope of a line is change in y over change in x, \({\Delta y \over \Delta x}\), and the y-intercept is the y-coordinate where the line crosses the vertical y-axis.
Solve for x:
-6x - 4 > -1 - 2x
| x > -5 | |
| x > 1\(\frac{1}{2}\) | |
| x > -\(\frac{3}{4}\) | |
| x > -\(\frac{5}{6}\) |
To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the > sign and the answer on the other.
-6x - 4 > -1 - 2x
-6x > -1 - 2x + 4
-6x + 2x > -1 + 4
-4x > 3
x > \( \frac{3}{-4} \)
x > -\(\frac{3}{4}\)
If the area of this square is 9, what is the length of one of the diagonals?
| \( \sqrt{2} \) | |
| 6\( \sqrt{2} \) | |
| 9\( \sqrt{2} \) | |
| 3\( \sqrt{2} \) |
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\( \sqrt{2} \)