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3 years ago

Solve the inequality. 2w 3 > 14 a. w 8.5 c. w 5.5

Mathematics

1 answer:

omeli [17]3 years ago

6 0

If there's a plus sign in between the 2w and 3 then the answer is C. w 5.5

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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

Use the given conditions to write an equation in point-slope form. Keep parenthesis in point-slope form. slope = 7, passing thro

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

y + 3 = 7(x - 1)

y + 3 = 7x - 7

y = 7x - 10

5 0

4 years ago

Find the derivative of the function at P 0 in the direction of A. f(x,y,z) = 3 e^x cos(yz), P0 (0, 0, 0), A = - i + 2 j + 3k

Alik [6]

The derivative of $f(x,y,z)$ at a point $p_0=(x_0,y_0,z_0)$ in the direction of a vector $\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k$ is

$\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}$

We have

$f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k$

and

$\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}$

Then the derivative at $p_0$ in the direction of $\vec a$ is

$3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}$

3 0

4 years ago

A mass oscillating up and down on the bottom of a spring can be modeled as harmonic motion. if they weight is displaced a maximu

snow_tiger [21]

9514 1404 393

Answer:

1.63 cm (across the centerline from release)

Step-by-step explanation:

If we assume time starts counting when we release the weight from its fully-extended downward position, then the position at 1.15 seconds can be found from ...

h(t) = -7cos(2πt/4)

h(1.15) = -7cos(π·1.15/2) = -7(-0.233445) ≈ 1.63412 . . . cm

That is, 1.15 seconds after the weight is released from below the resting position, it will be 1.63 cm above the resting position.

If it is released from above the resting position, it will be 1.63 cm below the resting position at t=1.15 seconds.

6 0

3 years ago

Asap! please help i’ll give brainliest

Alex777 [14]

well it would make sense to say that the longest fish is 12in longer than the shorter fish because if you take 21in (the longest fish) and subtract 9 (the shortest fish) you would get 12.

Hope that helped pls give brainlist

6 0

3 years ago