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lutik1710 [3]
3 years ago
15

The per-employee cost of insurance rose by 13% between 2015 and 2016. It then dropped by 2% between 2016 and 2017. It then incre

ased by 17% between 2017 and 2018. By what percentage did the cost of insurance change between 2015 and 2018? (NOTE: The answer is NOT 28%!)
Mathematics
1 answer:
Rama09 [41]3 years ago
5 0

Answer:

The cost of insurance change 29,6% between 2015 and 2018.

Step-by-step explanation:

We can start by defining C as the cost of insurance in 2015.

Then, we know it rose 13% between 2015 and 2016:

C_{16}=(1+0.13)*C=1.13C

It then dropped by 2% between 2016 and 2017:

C_{17}=(1-0.02)*C_{16}=0.98*(1.13C)=1.1074C

And then it increases by 17% between 2017 and 2018:

C_{18}=1.17C_{17}=1.17*1.1074C=1.295658C

We can conclude the cost of insurance change 29,6% between 2015 and 2018.

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HELP MEeeeeeeeee g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider<im
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(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for f:\mathbb R^2\to\mathbb R, we have the partial derivative with respect to x defined as

\displaystyle \frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y) - f(x,y)}h

The derivative at (0, 0) is then

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(0+h,0) - f(0,0)}h

• By definition of f, f(0,0)=0, so

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)}h = \lim_{h\to0}\frac{\tan^2(g(h,0))}{h\cdot g(h,0)}

• Expanding the tangent in terms of sine and cosine gives

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{h\cdot g(h,0) \cdot \cos^2(g(h,0))}

• Introduce a factor of g(h,0) in the numerator, then distribute the limit over the resulting product as

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{g(h,0)^2} \cdot \lim_{h\to0}\frac1{\cos^2(g(h,0))} \cdot \lim_{h\to0}\frac{g(h,0)}h

• The first limit is 1; recall that for a\neq0, we have

\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=1

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)}h = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h

and this is exactly the partial derivative of g with respect to x.

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h = \frac{\partial g}{\partial x}(0,0)

For the same reasons shown above,

\displaystyle \frac{\partial f}{\partial y}(0,0) = \frac{\partial g}{\partial y}(0,0)

(b) To show that f is differentiable at (0, 0), we first need to show that f is continuous.

• By definition of continuity, we need to show that

\left|f(x,y)-f(0,0)\right|

is very small, and that as we move the point (x,y) closer to the origin, f(x,y) converges to f(0,0).

We have

\left|f(x,y)-f(0,0)\right| = \left|\dfrac{\tan^2(g(x,y))}{g(x,y)}\right| \\\\ = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)^2}\cdot\dfrac{g(x,y)}{\cos^2(g(x,y))}\right| \\\\ = \left|\dfrac{\sin(g(x,y))}{g(x,y)}\right|^2 \cdot \dfrac{|g(x,y)|}{\cos^2(x,y)}

The first expression in the product is bounded above by 1, since |\sin(x)|\le|x| for all x. Then as (x,y) approaches the origin,

\displaystyle\lim_{(x,y)\to(0,0)}\frac{|g(x,y)|}{\cos^2(x,y)} = 0

So, f is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which f(x,y) changes as we move the point (x,y) closer to the origin, given by

\left|\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\right|,

approaches 0.

Just like before,

\left|\dfrac{\tan^2(g(x,y))}{g(x,y)\sqrt{x^2+y^2}}\right| = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)}\right|^2 \cdot \left|\dfrac{g(x,y)}{\cos^2(g(x,y))\sqrt{x^2+y^2}}\right| \\\\ \le \dfrac{|g(x,y)|}{\cos^2(g(x,y))\sqrt{x^2+y^2}}

and this converges to g(0,0)=0, since differentiability of g means

\displaystyle \lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)}{\sqrt{x^2+y^2}}=0

So, f is differentiable at (0, 0).

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