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2 years ago

The black graph is the graph of

Mathematics

1 answer:

Viktor [21]2 years ago

3 0

Step-by-step explanation:

the y values are still the same size as for the regular f(x).

so, answers c and d are out.

it is just that these same y values are happening in a more squeezed interval as for the regular f(x).

when we look at the endpoints for example, we find that

y = f(x) delivers y = -2 for x = 4

but the red function delivers y = -2 already for x = 2.

similar for y = 2, originally for x = 2, but in red for x = 1.

and for y = -1, originality for x = -2, but in red for x = -1.

let's call the original function black, and the new function red.

so, in other words red(2) = black(4)

red(1) = black(2)

red(-1) = black(-2)

red(x) = black(2x)

and therefore, a. is correct

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LiRa [457]

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3 years ago

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Suppose a change of coordinates T:R2→R2 from the uv-plane to the xy-plane is given by x=e−2ucos(5v), y=e−2usin(5v). Find the abs

anzhelika [568]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution is $\frac{\delta (x,y)}{\delta (u, v)} | = 10e^{-4u}$

Step-by-step explanation:

From the question we are told that

$x = e^{-2a} cos (5v)$

and $y = e^{-2a} sin(5v)$

Generally the absolute value of the determinant of the Jacobian for this change of coordinates is mathematically evaluated as

$| \frac{\delta (x,y)}{\delta (u, v)} | = | \ det \left[\begin{array}{ccc}{\frac{\delta x}{\delta u} }&{\frac{\delta x}{\delta v} }\\\frac{\delta y}{\delta u}&\frac{\delta y}{\delta v}\end{array}\right] |$

$= |\ det\ \left[\begin{array}{ccc}{-2e^{-2u} cos(5v)}&{-5e^{-2u} sin(5v)}\\{-2e^{-2u} sin(5v)}&{-2e^{-2u} cos(5v)}\end{array}\right] |$

$Let \ a = -2e^{-2u} cos(5v), \\ b=-2e^{-2u} sin(5v),\\c =-2e^{-2u} sin(5v),\\d=-2e^{-2u} cos(5v)$

$\frac{\delta (x,y)}{\delta (u, v)} | = |det \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] |$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | a * b - c* d |$

substituting for a, b, c,d

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | -10 (e^{-2u})^2 cos^2 (5v) - 10 e^{-4u} sin^2(5v)|$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | -10 e^{-4u} (cos^2 (5v) + sin^2 (5v))|$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = 10e^{-4u}$

7 0

3 years ago

Can u guys please answer these questions

aliya0001 [1]

Answer:

Im not sure about Q1 but I'll help with Q2

9 =

$\sqrt{81}$

8 =

$\sqrt[3]{512}$

14 =

$\sqrt{196}$

2 =

$\sqrt[3]{8}$

4 =

$\sqrt{16}$

I hope this helps :)

If you would like me to, i will respect your wishes and delete my answer.

8 0

2 years ago

I’m really struggling! Please help me!!

Andrew [12]

Answer:

8.0

Step-by-step explanation:

The triangle is a right triangle, so we are able to use trigonometric functions. Relative to the angle of 37°, we have the opposite side which is 6 and the adjacent side which is x. The trig function that uses the opposite and adjacent is tangent(SohCahToa). We can set up the following equation:

$tan(37) = \frac{opposite}{adjacent}$