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sergey [27]
3 years ago
11

Find 4 more than the quotient of 9 and 2

Mathematics
2 answers:
timurjin [86]3 years ago
8 0
The answer to the question would be 8.5
fenix001 [56]3 years ago
7 0
Im not sure but it should be 8.5
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Bob and Mark talk about their families. Bob says he has 3 kids, the product of their ages is 72. He gives another clue: the sum
Aloiza [94]
All there ages is 24 all together because if you divide 72/3 you get 24 then multiply it by 3 you will get 72
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What is pi for a cylinder​
grandymaker [24]

Answer:

Step-by-step explanation:

π Is a constant. It always has the same value.

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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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3 years ago
8(x + 7) - 12 = 0 can you solve the following inverse <br>​
vampirchik [111]

Answer:-11/2

Step-by-step explanation:

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2 years ago
Yesterday, I ran less than 5 kilometers but more than. 1,200 meters. How far could I have run?
taurus [48]
Less then 5 kilometers and more then 1,200 meters obv
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