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3 years ago

Complete the equation for f(x)

Mathematics

1 answer:

I am Lyosha [343]3 years ago

8 0

Answer:

f(x)=9

Step-by-step explanation:

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For each of the following vector fields F , decide whether it is conservative or not by computing the appropriate first order pa

Mazyrski [523]

Answer:

(a)

Conservative

(b)

Not conservative

(c)

Conservative.

Step-by-step explanation:

(a)

$\mathbf{F}(x,y) = (-10x+7y,7x+6y)$

Notice that

$\frac{\partial\mathbf{F}_y}{\partial x} = 7$

and

$\frac{\partial\mathbf{F}_x}{\partial y} = 7$

Therefore the field is conservative.

(b)

Notice that

$\mathbf{F}(x,y) = (-5y,-4x)$

and

$\frac{\partial\mathbf{F}_y}{\partial x} = -4$

but

$\frac{\partial\mathbf{F}_x}{\partial y} = -5$

Therefore is not conservative.

(c)

Notice that

To prove that the vector field is conservative you have to compute the curl of the vector field and you would get that.

$\mathbf{F}(x,y,z) = (-5x,-4y,1)$

$\nabla \times \mathbf{F} = (0,0,0)$

Therefore your vector field is conservative.

6 0

4 years ago

Which are equivalent (same value)? *<br> 31%<br> 31 per 100<br> 31/100<br> all of the above

Vinil7 [7]

Answer:

All of the above.

Step-by-step explanation:

31%=.31=31/100.

even if there is 31 per 100, it is the same. what if there is twice the value. 62/200. This is still reduced to the same number 31/100. There is still 31% of that thing.

5 0

3 years ago

Read 2 more answers

What Are Zeros of F(x)=3(×_5

Misha Larkins [42]

Answer:

The Zeros:

f(x)= 3(x –5)

X= 5

8 0

3 years ago

Verify that:

Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

$\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)$

Note that cot(x) = cos(x) / sin(x). Hence:

$\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)$

Multiply:

$\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

$\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Split:

$\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Simplify:

$\csc(x)-\sin(x)=\csc(x)-\sin(x)$

Problem 2)

We want to verify that:

$\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}$

Square:

$\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}$

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

$\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out the sin²(x) from the denominator:

$\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (perfect square trinomial):

$\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

$\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (difference of two squares):

$\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out a negative from the first factor in the denominator:

$\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Cancel:

$\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Distribute the negative into the numerator. Therefore:

$\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}$

3 0

3 years ago

I Just Wanna Be Nice And Gice Out A Free Brainliest :D

dolphi86 [110]

Answer:

THANK YOUUUU

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers