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MaRussiya [10]
3 years ago
12

C = 8πfb solve for f Formula to get (f) please :)

Mathematics
1 answer:
JulsSmile [24]3 years ago
5 0
C=8\pi fb\\
f=\dfrac{C}{8\pi b}
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[x÷(y–1)](−4)–[xy+(−3)]÷(−1) if x=−5, y=−2
IrinaK [193]
The result is 1/3.

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The first attempt in the attached picture has parentheses in the wrong place for the second term. This simplifies to
.. (-5/-3)*(-4) +(10 -3)
.. = -20/3 +7
.. = -(6 2/3) +7
.. = 1/3

5 0
3 years ago
A small publishing company is planning to publish a new book. The production costs will include one-time fixed costs (such as ed
Lady_Fox [76]

Answer:

for 3644 produced books the costs will be the same.

Step-by-step explanation:

x = number of books

f(x) = production method 1 costs

g(x) = production method 2 costs

f(x) = 11.75x + 57077

g(x) = 21.25x + 22459

for what number of books (x) will that be the same ?

11.75x + 57077 = 21.25x + 22459

34618 = 9.5x

x = 34618 / 9.5 = 3644

4 0
3 years ago
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Hoochie [10]
2,8 is the Awnser if you can see it on the graph it goes 2 up 8 right
5 0
3 years ago
An interior angle and an exterior angle are adjacent. The exterior angle measures 113°. What is the measure of the interior angl
Nuetrik [128]
Since all straight lines have an angle of 180 degrees if the exterior angle is equal to 113 degrees than you just subtract 113 from 180 which will equal 67.
5 0
3 years ago
Read 2 more answers
Solve the following equation:
Rama09 [41]

Complete the square.

z^4 + z^2 - i\sqrt 3 = \left(z^2 + \dfrac12\right)^2 - \dfrac14 - i\sqrt3 = 0

\left(z^2 + \dfrac12\right)^2 = \dfrac{1 + 4\sqrt3\,i}4

Use de Moivre's theorem to compute the square roots of the right side.

w = \dfrac{1 + 4\sqrt3\,i}4 = \dfrac74 \exp\left(i \tan^{-1}(4\sqrt3)\right)

\implies w^{1/2} = \pm \dfrac{\sqrt7}2 \exp\left(\dfrac i2 \tan^{-1}(4\sqrt3)\right) = \pm \dfrac{2+\sqrt3\,i}2

Now, taking square roots on both sides, we have

z^2 + \dfrac12 = \pm w^{1/2}

z^2 = \dfrac{1+\sqrt3\,i}2 \text{ or } z^2 = -\dfrac{3+\sqrt3\,i}2

Use de Moivre's theorem again to take square roots on both sides.

w_1 = \dfrac{1+\sqrt3\,i}2 = \exp\left(i\dfrac\pi3\right)

\implies z = {w_1}^{1/2} = \pm \exp\left(i\dfrac\pi6\right) = \boxed{\pm \dfrac{\sqrt3 + i}2}

w_2 = -\dfrac{3+\sqrt3\,i}2 = \sqrt3 \, \exp\left(-i \dfrac{5\pi}6\right)

\implies z = {w_2}^{1/2} = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i\dfrac{5\pi}{12}\right)}

3 0
2 years ago
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