Answer:
![g(x) = \sqrt[3]{x-1} - 2](https://tex.z-dn.net/?f=%20g%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx-1%7D%20-%202%20)
Step-by-step explanation:
We want to find h and k in:
![g(x) = \sqrt[3]{x-h} + k](https://tex.z-dn.net/?f=%20g%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx-h%7D%20%2B%20k%20)
At the inflection point, the second derivative is equal to zero, so:


Then x - h = 0.
Inflection point is located at (1, -2), replacing this x value we get:
1 - h = 0
h = 1
We know that the point (-2.5, -3.5) belongs to the function, so:
![-3.5 = \sqrt[3]{-2.5-1} + k](https://tex.z-dn.net/?f=%20-3.5%20%3D%20%5Csqrt%5B3%5D%7B-2.5-1%7D%20%2B%20k%20)
k ≈ -2
All data, used or not, are shown in the picture attached.
Answer:
B. 10 months
Step-by-step explanation:
The balance on the loan will be ...
b = 1600 - 80t . . . . . . where t is the number of months of payments
The balance in the savings account will be ...
s = 500 + 25t
The savings account balance will be at least as much as the loan balance when ...
s ≥ b
500 +25t ≥ 1600 -80t . . . substitute the account balance expressions
105t ≥ 1100 . . . . . . . . . . . . add 80t -500
t ≥ 1100/105 ≈ 10.48 ≈ 10
It will take Josh 10 months to have enough savings to pay the loan in full.
_____
<em>Comment on rounding</em>
IMO, it makes no sense to round down, as Josh will NOT have enough in 10 months. He will have enough after he makes one more payment of $80. At 10 months, the loan balance is $50 more than the savings balance. It will be 11 months before there is enough savings to pay off the loan.
Kyle, d = rt, right? This means that t = d/r. If we call the rate of the current c, then her rate upstream is hindered by the current and is 3 - c. Downstream is 3 + c. The total time is 6 hours... 5/(3 - c) + 5/(3 + c) = 6 Once we put everything over the same denominator, we don't need it anymore. 5(3 + c) + 5(3 - c) = 6(3 - c)(3 + c) 15 + 5c + 15 - 5c = 54 - 6c2
30 = 54 - 6c26c2 = 24c2 = 4c = 2 2 mph
Answer:
n-58
Step-by-step explanation:
$13 because Isabella will be awarded $104 making the ratio 1:8