Answer: 11 year
P(1) = 37,100
P(4) = 58,400
The linear equation (for x ≥ 1)
P(x) = 37,100 + a(x-1)
For x = 4
58,400 = 37,100 + a(4-1)
58,400 - 37,100 = 3a
21300 = 3a
a = 7100
So, the linear equation:
P(x) = 37100 + 7100*(x-1)
P(x) = 37100 + 7100x - 7,100
P(x) = 7100x + 30000
To find when the profit should reach 108100, we can substitute P(x) by 108100.
108100 = 7100x + 30000
108100 - 30000 = 7100x
78100 = 7100x
x = 78100/7100
x = 11
Answer: 11 year
The answer to the problem would be false
Answer:
15 or below
Step-by-step explanation:
10+10+8+8 = 36
51-36 = 15
Less than or equal to = 15 or below
distance between two points is [(x2-x1)^2 +(y2-y1)^2]^1/2
I.e,[{5-1)^2+(0-3)^2]^1/2
[16+9]^1/2
[25]^1/2
5
Answer:
a=1 or a=5
Step-by-step explanation:
+4=6a−1
Step 1: Subtract 6a-1 from both sides.
+4−(6a−1)=6a−1−(6a−1)
−6a+5=0
Step 2: Factor left side of equation.
(a−1)(a−5)=0
Step 3: Set factors equal to 0.
a−1=0 or a−5=0
a=1 or a=5
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