Manny buys 2 dozen of eggs how many did he buy in total?

LAST ONE I SWEAR BUT STILL WILL MARK BRAINLIEST

Svetlanka [38]

Actually I think it may be C

6 0

2 years ago

Read 2 more answers

John used AABC to write a proof of the Centroid Theorem. He began by drawing medians AK and CL,

kumpel [21]

Answer:

Hello your question is poorly written attached below is the complete question

answer : attached below

Step-by-step explanation:

To Prove: Z is located 2/3 of the distance from each vertex of ΔABC to the midpoint of the opposite side. we will apply ; property of bisecting a line , equality theorem , transitive property and similarity theorem

Attached below is the proof

8 0

2 years ago

Please Help!!<br><br>Use Euler’s formula to write in exponential form.

LekaFEV [45]

Answer:

C, $4e^{i(7\pi/4)}$

Step-by-step explanation:

To remind you, Euler's formula gives a link between trigonometric and exponential functions in a very profound way:

$e^{ix}=\cos{x}+i\sin{x}$

Given the complex number $2\sqrt{2}-2i\sqrt{2}$ , we want to try to get it in the same form as the right side of Euler's formula. As things are, though, we're unable to, and the reason for that has to do with the fact that both the sine and cosine functions are bound between the values 1 and -1, and 2√2 and -2√2 both lie outside that range.

One thing we could try would be to factor out a 2 to reduce both of those terms, giving us the expression $2(\sqrt{2}-i\sqrt{2})$

Still no good. √2 and -√2 are still greater than 1 and less than -1 respectively, so we'll have to reduce them a little more. With some clever thinking, you could factor out another 2, giving us the expression $4\left(\frac{\sqrt{2}}{2} -i\frac{\sqrt{2}}{2}\right)$ , and <em>now </em>we have something to work with.

Looking back at Euler's formula $e^{ix}=\cos{x}+i\sin{x}$ , we can map our expression inside the parentheses to the one on the right side of the formula, giving us $\cos{x}=\frac{\sqrt2}{2}$ and $\sin{x}=-\frac{\sqrt2}{2}$ , or equivalently:

$\cos^{-1}{\frac{\sqrt2}{2} }=\sin^{-1}-\frac{\sqrt2}{2} =x$

At this point, we can look at the unit circle (attached) to see the angle satisfying these two values for sine and cosine is 7π/4, so $x=\frac{7\pi}{4}$ , and we can finally replace our expression in parentheses with its exponential equivalent: