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3 years ago

Which of the following pairs of functions are inverses of each other?

Mathematics

1 answer:

crimeas [40]3 years ago

6 0

Answer:

i belive the answer is b but im not 100% sure

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Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

When y = 35,x = 2.5. If the value of y varies directly with x what is the value of y when the value of xis 3.25

leonid [27]

Answer:

The value of y would be 45.5

Step-by-step explanation:

To solve this problem, start with the base form of direct variation.

y = kx

Now we can use our original values to model the equation and find k.

35 = k(2.5)

14 = k

Now we can model the equation as:

y = 14x

Now to find y, when x = 3.25, simply put 3.25 into the equation.

y = 14(3.25)

y = 45.5

8 0

3 years ago

Can someone help please

xxMikexx [17]

D- It is an enlargement with a scale factor greater than 1.

From the pre-image to the image, the scale factor is 3.

5 0

3 years ago

Write the expression as one involving only sin x and cos x. sin 2x + cos x

AfilCa [17]

Answer:

sin 2x + cos x = 2 sin x cos x + cos x = (2 sin x + 1)cos x

Step-by-step explanation:

Given the expression: sin 2x + cos x,

then we can use the formula: sin 2x = 2 sin x cos x, which gives:

sin 2x + cos x = 2 sin x cos x + cos x = (2 sin x + 1)cos x

So there you have two expressions in terms of sin x and cos x, as requested. :D

6 0

1 year ago

A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down t

VLD [36.1K]

Answer:

6000 ft

Step-by-step explanation:

Let length of rectangular field=x

Breadth of rectangular field=y

Area of rectangular field= $1500000$ square ft

Area of rectangular field= $l\times b$

Area of rectangular field= $x\time y$

$1500000=xy$

$y=\frac{1500000}{x}$

Fencing used ,P(x)= $x+x+y+y+y=2x+3y$

Substitute the value of y

P(x)= $2x+3(\frac{1500000}{x})$

$P(x)=2x+\frac{4500000}{x}$

Differentiate w.r.t x

$P'(x)=2-\frac{4500000}{x^2}$

Using formula: $\frac{dx^n}{dx}=nx^{n-1}$

$P'(x)=0$

$2-\frac{4500000}{x^2}=0$

$\frac{4500000}{x^2}=2$

$x^2=\frac{4500000}{2}=2250000$

$x=\sqrt{2250000}=1500$

It is always positive because length is always positive.

Again differentiate w.r.t x

$P''(x)=\frac{9000000}{x^3}$

Substitute x=1500

$P''(1500)=\frac{9000000}{(1500)^3}>0$

Hence, fencing is minimum at x=1 500

Substitute x=1 500

$y=\frac{1500000}{1500}=1000$

Length of rectangular field=1500 ft

Breadth of rectangular field=1000 ft

Substitute the values

Shortest length of fence used= $2(1500)+3(1000)=6000 ft$

Hence, the shortest length of fence that the rancher can used=6000 ft

3 0

3 years ago

Which of the following pairs of functions are inverses of each other?​

Which of the following pairs of functions are inverses of each other?