This question is incomplete
Complete Question
In the first half of a basketball game, a player scored 9 points on free throws and then scored a number of 2-point shots. In the second half, the player scored the same number of 3-point shots as the number of 2-point shots scored in the first half. Which expression represents the total number of points the player scored in the game?
a) 2x + 3x + 9
b) 2x + 3 + 9
c) 2x + 3x + 9x
d) 2 + 3x + 9
Answer:
a) 2x + 3x + 9
Step-by-step explanation:
Let the number of points shots a player scores = x
In a free throw, the player scored 9 points = 9
The player also scored a number of 2-point shots = 2x
In the second half, the player scored the same number of 3-point shots as the number of 2-point shots scored in the first half = 3x
The expression represents the total number of points the player scored in the game =
2x + 3x + 9
Answer:
The answer to this is fa
Step-by-step explanation:
This is because you as the solver doesn't know what the variables f or a is so you put them together as multiplication until you do understand what f and a are so you can multiply them.
If you were to graph this it would be a horizontal line going infinite in both directions. That means no slope or y axis.
Standard form Ax + By = C
so
<span>(y - y1) = m(x - x1)
</span>given (5, -7) and m = -1/5
so
y + 7 = -1/5(x - 5)
-5y - 35 = x - 5
x + 5y = -30
answer
d. x + 5y = -30
Answer:
to find a GCF you need 2 numbers. You only have one here
Step-by-step explanation:
Answer:
Outside the circle
Step-by-step explanation:
Let's first write the equation of this circle: 
, where (h, k) is the center and r is the radius. Here, the center is (-6, -2). We need to find the radius, which will just be the distance from N to E:
NE = 
The radius is √34, which means that r² = 34. So, our equation is:
(x + 6)² + (y + 2)² = 34
Plug in -10 for x and -7 for y:
(x + 6)² + (y + 2)² = 34
x² + 12x + 36 + y² + 4y + 4 = 34
x² + 12x + y² + 4y + 40 = 34
x² + 12x + y² + 4y + 6 = 0
(-10)² + 12 * (-10) + (-7)² + 4 * (-7) + 6 = 7
Since 7 > 0, we know that H lies outside the circle.
