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2 years ago

This graph shows a portion of an odd function.

Mathematics

1 answer:

nexus9112 [7]2 years ago

6 0

9514 1404 393

Answer:

-2
-1
-2
-3

Step-by-step explanation:

An odd function is symmetrical about the origin:

f(-x) = -f(x)

So, ...

f(-2) = -f(2) = -2

f(-3) = -f(3) = -1

f(-4) = -f(4) = -2

f(-6) = -f(6) = -3

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Sally plans to enclose a square garden with fencing. The area of the garden is 1600 square feet. How many feet of fencing does s

Llana [10]

Answer:

160 ft

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

Consider F and C below. F(x, y, z) = y2 sin(z) i + 2xy sin(z) j + xy2 cos(z) k C: r(t) = t2 i + sin(t) j + t k, 0 ≤ t ≤ π (a) Fi

Liono4ka [1.6K]

Answer:

a) $f (x,y,z)= xy^2\sin(z)$

b) $\int_C F \cdot dr =0$

Step-by-step explanation:

Recall that given a function f(x,y,z) then $\nabla f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})$ . To find f, we will assume it exists and then we will find its form by integration.

First assume that F = $\nabla f$ . This implies that

$\frac{\partial f}{\partial x} = y^2\sin(z)$ if we integrate with respect to x we get that

$f(x,y,z) = xy^2\sin(z) + g(y,z)$ for some function g(y,z). If we take the derivative of this equation with respect to y, we get

$\frac{\partial f}{\partial y} = 2xy\sin(z) + \frac{\partial g}{\partial y}$

This must be equal to the second component of F. Then

$2xy\sin(z) + \frac{\partial g}{\partial y}=2xy\sin(z)$

This implies that $\frac{\partial g}{\partial y}=0$ , which means that g depends on z only. So $f(x,y,z) = xy^2\sin(z) + g(z)$

Taking the derivative with respect to z and making it equal to the third component of F, we get

$xy^2\cos(z)+\frac{dg}{dz} = xy^2\cos(z)$

which implies that $\frac{dg}{dz}=0$ which means that g(z) = K, where K is a constant. So

$f (x,y,z)= xy^2\sin(z)$

b) To evaluate $\int_C F \cdot dr$ we can evaluate it by using f. We can calculate the value of f at the initial and final point of C and the subtract them as follows.