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3 years ago

Plot the points (2,5 )and (2,-7)

Mathematics

2 answers:

Allisa [31]3 years ago

4 0

Step-by-step explanation:

Go over to two on the x axis they go up to five to plot (2,5 ). To plot (2,-7) go over to the the two then go up to -7.

WINSTONCH [101]3 years ago

3 0

Answer:

It's in the 3rd collum

Step-by-step explanation:

You might be interested in

How do I create a histogram for the data?

Andrews [41]

Answer:

Group by buckets and use that information to make your histogram.

Step-by-step explanation:

When you make a histogram you need to group elements into buckets, and out of those buckets make your histograms. You could for example take the mean or the median of groups and out of that information make the histogram. That is your decision.

For example that you have

JM 85

JM 73

so the mean between 85 and 73 is 79 , so in your histogram the height of JM would be 79.

To summarize, you group by buckets and use that information to make your histogram.

4 0

4 years ago

Help please!!!!!!!!!!!

ss7ja [257]

First, solve 7 times square root of 5.

7 x 2.236 = 15.652

Now find the solution above that results in 15.65.

We can quickly eliminate the top 2 as they will result in very large numbers.

The bottom problem is quick to check - simplifies to (7 + 5)/5 = 2.4. No

It looks like the next to last answer is our best bet.

Figure the square root of 196 (14) and multiply by square root of 5 (2.236). Giving us 31.304.

Divide the result by 2. And that gives us 15.652.

If you knew the square root of 196 was 14, it was helpful to find this quickly because 14/2 = 7 and then the other square root was the 5 from the problem.

5 0

3 years ago

Z^4-5(1+2i)z^2+24-10i=0

mixer [17]

Using the quadratic formula, we solve for $z^2$ .

$z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2$

Taking square roots on both sides, we end up with

$z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}$

Compute the square roots of -171 + 140i.

$|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221$

$\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)$

By de Moivre's theorem,

$\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i$

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

$t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}$

as well as the fact that

$0$

$\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}$

(whose signs are positive because of the domain of $\frac t2$ ).

This leaves us with

$z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}$

Compute the square roots of 5 + 12i.

$|5 + 12i| = \sqrt{5^2 + 12^2} = 13$

$\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)$

By de Moivre,

$\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i$