All categories

3 years ago

Calculate the distance between (−3 + 4i) and (3 −3i) on the coordinate plane.

1 answer:

7 0

Another one for my "impossible math problems" file.

The distance between real parts of these numbers is 3-(-3) = 6.
The distance between imaginary parts of these numbers is (4-(-3)) = 7.
The straight-line distance on the complex plane between these numbers is given by the Pythagorean theorem as
distance = √(6² + 7²)
distance = √85 ≈ 9.2195

_____
If your answer choices are supposed to have radical signs, then the best choice is ...
D) √85

Otherwise, no offered choice is correct.

You might be interested in

A salesperson's weekly paycheck is 10% more than a second salesperson's paycheck. The two paychecks total $1075. Find the amount

Ksivusya [100]

Originally without the 10% extra, both have 100%. since the question stated that the first salesperson have 10% more, the percentage of his paycheck will be a total of 110%. on the other hand, the second salesperson will have only 100%. the question then stated that when both oaychecks are added together, it will give a total of $1075.

110% + 100% represents $1075
210% represents $1075

so now you need to find what is 1 %
1% represents $(1075/210)

the first salesperson have 110%, so you
(1075/210) × 110% = $563. 0952381
round off to the nearest cent so it will become $563.10
the first salesperson paycheck is $563.10

the second salesperson paycheck 100%. so you....
(1075/210) × 100% = $511.9047619
=$511.90
therefore the second salesperson paycheck is $511.90.

4 0

3 years ago

The area of the rectangle is 24 inches. How much longer is its length than its width?

antiseptic1488 [7]

I Belive That There Can't Be 3 Numbers In A Rectangle Because At The Top You Get 24 And 6 And I Think It Is BxH One Half Times Base Times Height So 2xBxH .

8 0

3 years ago

226,710 - 724,435 =<br> How to work this out without calculator

Anit [1.1K]

Okay so this is going to be confusing but work from right to left. since you can’t do 0-5, borrow from the one next to the zero on the top, cross out the one and put zero and add a one on top of the zero which makes it 10. now you can do 10-5 to get 5. put the five down below the line. next solve the second column of numbers ( second to the right). since you cannot do 0- 3, like the last one, you have to borrow from the 7 over to the left. take one from there and cross the seven out and put a six. now you have 10-3 which is 7. put the 7 down. now do move to the next column and do 6-2 to get 4. put the four down below the line. move to the next column and do 2-2 to get 0. put the zero down below the line. lastly, we know the answer has to be negative so we are going to do 2-7 to get -5. put -5 down below the line and together to get the answer -502,275. i know that was very confusing but i hope it helped a little bit. i added a picture to clarify a little

7 0

2 years ago

Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

leonid [27]

Let $P=(x,y,z)$ be an arbitrary point on the surface. The distance between $P$ and the given point $(7,11,0)$ is given by the function

$d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}$

Note that $f(x)$ and $f(x)^2$ attain their extrema, if they have any, at the same values of $x$ . This allows us to consider the modified distance function,

$d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2$

So now you're minimizing $d^*(x,y,z)$ subject to the constraint $z^2=xy$ . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

$\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)$

which has partial derivatives

$\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}$

Setting all four equation equal to 0, you find from the third equation that either $z=0$ or $\lambda=-1$ . In the first case, you arrive at a possible critical point of $(0,0,0)$ . In the second, plugging $\lambda=-1$ into the first two equations gives

$\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10$

and plugging these into the last equation gives

$z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5$

So you have three potential points to check: $(0,0,0)$ , $(2,10,2\sqrt5)$ , and $(2,10,-2\sqrt5)$ . Evaluating either distance function (I use $d^*$ ), you find that

$d^*(0,0,0)=170$
$d^*(2,10,2\sqrt5)=46$
$d^*(2,10,-2\sqrt5)=46$

So the two points on the surface $z^2=xy$ closest to the point $(7,11,0)$ are $(2,10,\pm2\sqrt5)$ .

5 0

3 years ago

Cassie is a half pound of sugar in her cabinet her cake recipe calls for two tenths of a pound of sugar how many cakes can she m

mamaluj [8]

The information's given in the question should be very minutely checked. Based on those given information's the answer to the question can easily be deduced.
Amount of sugar in the cabinet of Cassie = 1/2 pound
Amount of sugar required by Cassie to make her cake recipe = 2/10 pound
   = 1/5 pounds
Then
The number of cakes that Cassie can make = (1/2)/(1/5)
      = 5/2
      = 2 1/2
So from the above deduction we can easily conclude that Cassie can make a total of 2 cakes with the sugar she has and there will still be some sugar left.

7 0

3 years ago