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3 years ago

Helply help help help

Mathematics

1 answer:

Rainbow [258]3 years ago

6 0

It is D parabola because in a double right circular cone the vertex is in the middle of two opposite cones and cutting through the vertex is making parabola.

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P-4=-9+p

pantera1 [17]

Answer:

no solution

Step-by-step explanation:

p-4=-9+p

p=-5+p

0=-5

8 0

3 years ago

Match each table with the equation that represents the same proportional relationship. Match each table with the equation that r

Alisiya [41]

29 tables of relationships is the answer to your equation above … THANK ME LATER

7 0

2 years ago

What is the exact area of a circle with a diameter of 10 cm? a. 5πcm^2 b. 10πm^2 c. 25πcm^2 d. 100πcm^2

dalvyx [7]

Answer:

Step-by-step explanation:

a= $\pi$ r^2

To find the radius, divide the diameter by 2

r=d/2

r=10/2

r=5

a= $\pi$ r^2

a= $\pi$ 5^2

a=25 $\pi$

So, choice C is correct

Hope this helps! :)

4 0

3 years ago

9 x 10' =_________________________________

elena-14-01-66 [18.8K]

Answer:

Uh 90 i hope your kidding

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

3 0

2 years ago