1. Find the slope of the line that passes through (-2,1), (1,10).
a. 3
b. -3
c. 1/3
d. -1/3
2. Which equation represents a direct variation? What is the constant variation?
a.3y = 2x + 1; 1
b.y = -5x - 11; 5
c.4y = -12x; -3
d.y + 7 = 2x - 1; 7
3. Suppose y varies directly with x, and y = 12 when x = -3. What is the value of y when x = 6.
a. -24
b. 2
c. -2
d. -4
4. What is the equation of a line with a slope of -7 and a y-intercept of 6?
a.6y = -7x
b.y = 6x + (-7)
c.-7y = x + 6
d.y = -7x + 6
5. What is the equation of the line through (5,1) with a slope of -3?
a.y + 1 = 3(x + 5)
b.y - 1 = 3(x - 5)
c.y - 1 = -3(x - 5)
d.y + 1 = -3(x - 5)
7 years ago
5 years ago
1. Find the slope of the line. (1 point)
2
β2
Β½
-Β½
2. Find the slope of the line that passes through (β2, 1), (1, 10). (1 point)
3
β3
β
-β
3. Which equation represents a direct variation? What is the constant of variation? (1 point)
3y = 2x + 1; 1
y = β5x β 11; 5
4y = β12x; β3
y + 7 = 2x β 1; 7
4. Suppose y varies directly with x, and y = 12 when x = β3. What is the value of y when x = 6? (1 point)
β24
2
β2
β4
5. What is the equation of a line with a slope of β7 and a y-intercept of 6? (1 point)
6y = β7x
y = 6x + (β7)
β7y = x + 6
y = β7x + 6
6. What is the equation of the line that passes through the points (0, 4) and (3, β2)? (1 point)
y = 2x + 0
y = β 1/2x + (β2)
y = β2x + 4
y = 1/2x+ 4
7. What is the equation of the line through (5, 1) with a slope of β3? (1 point)
y + 1 = 3(x + 5)
y β 1 = 3(x β 5)
y β 1 = β3(x β 5)
y + 1 = β3(x β 5)
8. What is the equation of the given line in point-slope form? (1 point)
y β 1 = 2(x + 2)
y β 1 = 2(x β 2)
y β 2 = 2(x + 1)
y β 2 = (x + 2)
2 years ago
So whatβs the answer?
2 years ago
anyone got the answers yet
2 years ago
Answers for mid-unit review practice are
1. B
2. A
3. C
4. A
5. D
6. C
7. C
8. A
2 years ago
>.< is completely correct!
11 months ago
1. To find the slope of a line passing through two points, we can use the formula:
slope = (y2 - y1)/(x2 - x1)
In this case, the points are (-2,1) and (1,10). Plugging these values into the formula, we get:
slope = (10 - 1)/(1 - (-2))
= 9/3
= 3
Therefore, the slope of the line is 3.
Answer: a. 3
2. In a direct variation, the equation is of the form y = kx, where k is the constant of variation.
Looking at the given options, we can see that the equation c. 4y = -12x represents a direct variation, since it can be rewritten as y = -3x. In this equation, the constant of variation is -3.
Answer: c. 4y = -12x; -3
3. In a direct variation, if we know the value of y when x is a certain value, we can find the value of y when x is a different value by using the equation y = kx, where k is the constant of variation.
In this case, we are given that y = 12 when x = -3. To find the value of y when x = 6, we can use the equation:
y = kx
Substituting the given values, we get:
12 = k(-3)
Solving for k, we get:
k = -4
Now, we can find the value of y when x = 6:
y = (-4)(6)
= -24
Therefore, the value of y when x = 6 is -24.
Answer: a. -24
4. The equation of a line can be written in the form y = mx + b, where m is the slope and b is the y-intercept.
In this case, the given slope is -7 and the y-intercept is 6. Plugging these values into the equation, we get:
y = (-7)x + 6
Therefore, the equation of the line is y = -7x + 6.
Answer: d. y = -7x + 6
5. To find the equation of a line given a point (x,y) and the slope, we can use the point-slope form of the equation:
y - y1 = m(x - x1)
In this case, the given point is (5,1) and the slope is -3. Plugging these values into the equation, we get:
y - 1 = -3(x - 5)
Expanding and simplifying, we get:
y - 1 = -3x + 15
Rearranging to the standard form of the equation, we get:
y = -3x + 16
Therefore, the equation of the line through (5,1) with a slope of -3 is y = -3x + 16.
Answer: None of the given options.