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3 years ago

I need help with my math problems

Mathematics

2 answers:

Oxana [17]3 years ago

7 0

13 16 15 19 20 15 19 20 14 15 17 18 and that's all i see hopes that helped :)

makvit [3.9K]3 years ago

6 0

13, 16, 15, 19,20,15,19,20,14,15,17,19,

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Write the point-slope equation of (1,6) (3,-6)

dangina [55]

Answer:

0 (I think)

Step-by-step explanation:

(1,6) (3,-6)

X1 Y1. X2 Y2

-6-6= 0

3-1=2

0/2= 0

4 0

3 years ago

Read 2 more answers

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).

3 0

3 years ago

What is the missing dimension of a rectangle or prism with the following dimensions and volume? Length 3 cm width 5 cm volume 45

olasank [31]

Hello!

In order to find out the volume of any prism, you are to do length x width x height.

Because we only know the length and the width, we can multiply those to get started. 3 x 5 is 15, so we know that's what we can start with.

Because we know the volume, all we have to do is do h x 15 = volume (45). Now, what is height?

In order to find the height, we must find a number that, when multiplied by 15, equals 45.

So we can try out 3, which when multiplied by 15, equals 45.

That means 3 is your height.

Hope I helped! :)

3 0

2 years ago

Choose the equation below that represents the line passing through the point (-3,-1) with a slope of 4.

Bingel [31]

Y = mx + b
slope(m) = 4
(-3,-1)...x = -3 and y = -1
now we sub...we r looking for b, the y int.
-1 = 4(-3) + b
-1 = -12 + b
-1 + 12 = b
11 = b

so ur equation is : y = 4x + 11

8 0

2 years ago

Simplify the expression.<br> - (m<br> m + 3n - 13)

natulia [17]

Answer:

$= - {m}^{2} - 3n + 13$