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3 years ago

a. Miss Butler made a circle graph to show the grades on the term papers. There were 14 B's, 10 A's, and 8 C's.

Mathematics

1 answer:

dexar [7]3 years ago

5 0

I actually I’m really confused sorry can’t answer

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3/8 + 1/4 + 1/2 - 2/3 =

Nostrana [21]

Answer:

$\frac{11}{24}$

Step-by-step explanation:

3/8 + 1/4 + 1/2 - 2/3

- > 1/4 = 2/8

3/8 + 2/8 + 1/2 - 2/3

5/8 + 1/2 - 2/3

- > 1/2 = 4/8

5/8 + 4/8 - 2/3

9/8 - 2/3

- > LCM of 8,3: 24

- > 9/8 = 27/24

- > 2/3 = 16/24

27/24 - 16/24

11/24

Hope this helps you.

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3 years ago

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Thirty three thousand and ninety one

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Rohan buys 9 tangerines for the same price each, and 1 watermelon for $4. He spends less than $14 altogether. Which inequality c

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9x + 4 < 14 is the required inequality.

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How to find volume of the liquid

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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

1 year ago

a. Miss Butler made a circle graph to show the grades on the term papers. There were 14 B's, 10 A's, and 8 C's.​

a. Miss Butler made a circle graph to show the grades on the term papers. There were 14 B's, 10 A's, and 8 C's.