Answer:
Linear equation with a slope of 2 that goes through the point (3, 4) is 
.
Step-by-step explanation:
From statement we know the slope of the line and a point contained in it. Using the slope-point equation of the line is the quickest approach to determine the appropriate equation, whose expression is:
Where:
- Slope, dimensionless.
, 
- Components of given point, dimensionless.
, 
- Independent and dependent variable, dimensionless.
If we know that 
, 
and 
, the linear equation is found after algebraic handling:
1) 
Given
2) 
Compatibility with Addition/Existence of Additive Inverse/Modulative Property
3) Distributive Property/
/Definition of sum/Result
Linear equation with a slope of 2 that goes through the point (3, 4) is .
Answer:
P(seventh grade winner) = 3/8
Step-by-step explanation:
If you add a value to the outcome of the function (as in case a), you will just shift the graph up along the y-axis. Makes sense, right, since you're adding an extra 2 to the value you will plot on the y axis.
If you subtract (in this example) a value to the input of the function, you're actually "looking back" into the past of the graph (i.e., the left side). For example, for x=4 you're now plotting the value that used to be at x=1. So this shifts the graph to the right.
That may be confusing at first. But I hope with a little thought experiment, you get it.
B. an internal retaining ring
hope this is the correct one
Answer:
12). LM = 37.1 units
13). c = 4.6 mi
Step-by-step explanation:
12). LM² = 23² + 20² - 2(23)(20)cos(119)°
LM² = 529 + 400 - 920cos(119)°
LM² = 929 - 920cos(119)°
LM = 
= 
= 37.08
≈ 37.1 units
13). c² = 5.4² + 3.6² - 2(5.4)(3.6)cos(58)°
c² = 29.16 + 12.96 - 38.88cos(58)°
c² = 42.12 - 38.88cos(58)°
c = 
c = 
c = 4.6386
c ≈ 4.6 mi
