All categories

3 years ago

Can someone help me with this proof

Mathematics

2 answers:

larisa86 [58]3 years ago

8 0

The fourth option is the correct answer.

Hope this helps! ;)

mezya [45]3 years ago

3 0

The fourth option hoped it helped

You might be interested in

easy algebra question below first correct answer gets brainliest, if you put a link i will report and block

boyakko [2]

Answer:

6x²

Step-by-step explanation:

3x×2x=6...... then an x²

Therefore answer is 6x²

3 0

3 years ago

Simplify each expression.

lutik1710 [3]

Answer:

1. 80

2. 18.9

3. 7

4. 146

5. 2

6. 4

Step-by-step explanation:

1. 6+4=10

10×8=80

2. 3.6×4.5= 16.2

16.2+2.7= 18.9

3. 3+2=5

5×2= 10

17-10=7

4. 4×3=12^2

12×12=144

144+2= 146

5. 3+2=5

5×6=30

30÷15=2

6. 5×2= 10

40÷10=4

4 0

3 years ago

The table represents the start of the division of 8x^4+2x^3-7x^2+3x-2 Find the quotient.

xxTIMURxx [149]

Answer:

idkidkidkiskidkisdusdwasd

7 0

3 years ago

What is the fill in the blank answers?

Olenka [21]

Think first is Distributive Property
Combine any LIKE TERMS
Isolate the VARIABLE
By using INVERSE
Check your ANSWER
I think these are the answers hope it helps!

4 0

3 years ago

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

3 years ago