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2 years ago

A right pyramid with a regular hexagon base has a height of 3 units.

Mathematics

1 answer:

mina [271]2 years ago

8 0

Answer: in order, the answer is 3, height of the pyramid, 6, then 108

Step-by-step explanation: I JUST DID THE QUESTION

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5. A large firm hires 3% more workers which brings

Natasha2012 [34]

Answer:

425733.32

Step-by-step explanation:

3%= 12772×100÷x

3x= 12772×100

x=425733.34

5 0

2 years ago

The left and right page numbers of an open book are two consecutive integers whose sum is 229

alukav5142 [94]

Answer:

114, and 115

Step-by-step explanation:

114+115 = 229

6 0

2 years ago

Read 2 more answers

Need help with gem trip sequences <br> 36 points!!!!!!!!!

coldgirl [10]

Answer:

$a_{n}$ = 8 $(-3)^{n-1}$

Step-by-step explanation:

The n th term of a geometric sequence is

$a_{n}$ = a $(r)^{n-1}$

where a is the first term and r the common ratio

Here a = 8 and r = - 24 ÷ 8 = - 3, thus

$a_{n}$ = 8 $(-3)^{n-1}$

3 0

3 years ago

The value of Maggie’s car decreased by 20% since last year when she bought it if the car is not worth 21,000 how much was the ca

leonid [27]

Answer:

Step-by-step explanation:

You would do 21,000×0.20, which equals 4,200. Since 4,200 equals 20% of 21,000, then you would do 21,000-4,200, which equals 16,800. So the car is worth $16,800 dollars this year.

6 0

2 years ago

Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

2 years ago