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3 years ago

13

What’s 1440 feet below sea level

1 answer:

astra-53 [7]3 years ago

7 0

-1440 feet below sea level in integer form

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Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Alexxx [7]

Hi There!

Answer: Bowler 2

Why: if you notice that line in the box plot shows the median and it specifically shows that the meadian is 20 or so greater than Bowler 1.

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4 years ago

Read 2 more answers

The taxi cost $2.25 for the first mile plus $.15 Cents for each additional tenth of a mile. For a 5.2-mile trip, Eric paid $10 a

Readme [11.4K]

The driver's tip was $1.45 because the trip cost $8.55 (2.25+1.5+1.5+1.5+1.5+ 0.3) and Eric gave the driver $10 so to find how much the driver got tipped, you need to subtract 8.55 from 10 so $10 - $ 8.55 = $1.45

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3 years ago

What makes two angles exterior versus interior ?<br>

astraxan [27]

Answer:

Any angle that is on the inside of a shape is called an interior angle.

Step-by-step explanation:

Any angle that is on the outside of a shape is called an exterior angle. An exterior angle is made by extending one of the lines of the shape beyond the intersection point. ... Notice that the two angles make a straight line.

4 0

3 years ago

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

PLease help due today! I will give brainliest!!

Roman55 [17]

The answer is A
Hopes this helps

6 0

2 years ago