To solve for n start off by adding bn to both sides.
an - 4a + 3 + -bn = bn + -bn
an - bn - 4a + 3 = 0
Now, add -3 to both sides.
an - bn - 4a + 3 + -3 = 0 + -3
an - bn - 4a = -3
Add 4a to both sides.
an - bn - 4a + 4a = -3 + 4a
an - bn = 4a - 3
Factor out variable n.
n(a-b) = 4a - 3
Divide both sides by a - b.
a(a-b) / a - b = 4a - 3 / a - b
Your final answer is:
n = 4a - 3 / a - b
Answer:
1) x = 6, y = -2 + 1/3x
2)
Step-by-step explanation:
1) x-3y=6
-3y=6-x, 6-x/-3y
y = -2 + 1/3x
x - 3(-2+1/3x) = 6
x - 6 + x = 6
2x =12
x = 6
2) x=3y+4
-3y = -x+4
y = -x/-3 +4/-3
y = 1/3x + -4/3
x = 3(1/3x + -4/3)
I am unsure about x on number two...
Answer:
309 meters per second
Step-by-step explanation:
Answer:
The unit price is the cost per unit of an item or the cost/price for each item.
1) 4$ per pound. By simplifying the proportion (constant ratio) between the cost, and the pounds of apples. 3 pounds of apples cost 12$ → 3/3 pounds of apples cost 12/3$ → 4 dollars for every pound.
2) 2$ per pound. By evaluating the rate of change (change in the y over x or dependent variable over independent) in the equation: y = 2x. y is the cost in dollars, and x is the pounds of apples. So there are 2 pounds (weight) of apples for every dollar.
3) 3$ per pound. Given a graph with a y scaled by 3, and an x scaled by 1 with a graph y = x or 1 unit up for every unit right. This must be equivalent to y = 3x. Where y is labeled as the cost in dollars, and x as the weight in pounds. So there are 3 dollars for every pound of apples.
4) Store B. Because 2 is less than 3 which is less than 4.
Step-by-step explanation:
Since they want to prove that ABCD is a square, you only need to know the lengths of two adjacent sides.
Distance formula: d = 
(x2 - x1)^2 + (y2 - y1)^2
A: (3, 4) B: (2, -2) C: (-4, -1) D: (-3, 5)
Distance of AB where A = (x1, y1) and B = (x2, y2)
d = (2 - 3)^2 + (-2 - 4)^2
d = (-1)^2 + (-6)^2
d = 1 + 12
d = 13
Distance of CB where B = (x1, y1) and C = (x2, y2)
d = (-4 - 2)^2 + (-1 - (-2))^2
d = (-6)^2 + (1)^2
d = 12 + 1
d = 13
Since dAB = dCB, therefore quadrilateral ABCD is a square.
