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Andrews [41]
3 years ago
7

The formula i = prt gives a simple interest \ earned on an account where the amount is p is deposited at an interest rate r for

a certain number of years t. Use the table to order the accounts from least to greatest interest earned after 5 years.
Mathematics
1 answer:
jeyben [28]3 years ago
8 0
In order to answer this we need the table, however to do this all you would do would be to multiply the p value in the table with the corresponding r value in the table and then multiply this by 5, then order them, smallest to greatest.
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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
GrogVix [38]

(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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Answer:

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Answer:

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Step-by-step explanation:

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Answer:

-14.53

Step-by-step explanation:

-5.23-9.3=-14.53 it's a negative plus a negative

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What is 2x + 3y -4 =? <br><br> x = 2 , y =4
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