All categories

3 years ago

Please help with this i would be so glad and happy.

Mathematics

1 answer:

weqwewe [10]3 years ago

3 0

Answer:

Step-by-step explanation:

d) (-2 +4i) -( 3 + 9i) = -2 + 4i - 3 - 9i

= -2 - 3 + 4i - 9i

= -5 - 5i

e)3i(6 - 5i) = 3i *6 - 5i * 3i

= 18i - 15*i² {i² = -1}

= 18i -15*(-1)

= 18i + 15

= 15 + 18i

f) (-3 + 7i)(1 - 2i) = (-3)*1 - 2i * (-3) + 7i *1 - 2i* 7i

= -3 + 6i + 7i - 14i²

= -3 + 6i + 7i - 14*(-1)

= -3 + 6i+ 7i + 14

= 11 + 13i

You might be interested in

What is n in n:44 = 7:11?

Lady bird [3.3K]

Answer:

n = 28

Step-by-step explanation:

hope this helps :)

8 0

3 years ago

Read 2 more answers

42×10 minutes in shower

ludmilkaskok [199]

42 times 10 is 420. hope this helps

6 0

3 years ago

Read 2 more answers

Find the taylor series for f(x centered at the given value of

Lady_Fox [76]

To find the Taylor series for f(x) = ln(x) centering at 5, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

$f(x) = ln(x) \\ f^{1}(x) = \frac{1}{x} \\ f^{2}(x) = -\frac{1}{x^{2}} \\ f^{3}(x) = \frac{2}{x^{3}} \\ f^{4}(x) = \frac{-6}{x^{4}}$
.
.
.
Since we need to have it centered at 5, we must take the value of f(5), and so on.

$f(5) = ln(5) \\ f^{1}(5) = \frac{1}{5} \\ f^{2}(5) = \frac{-1}{5^{2}} \\ f^{3}(5) = \frac{1(2)}{5^{3}} \\ f^{4}(5) = \frac{-1(2)(3)}{5^{4}}$
.
.
.
Following the pattern, we can see that for $f^{n}(x)$ ,

$f^{n}(x) = (-1)^{n-1} \frac{1(2)(3)...(n-1)}{5^{n}} \\ f^{n}(x) = (-1)^{n-1} \frac{(n-1)!}{5^{n}}$

This applies for $n\geq 1$ . Expressing f(x) in summation, we have

$\sum_{n=0}^{\infty} \frac{f^{n}(5)}{n!} (x-5)^{n}$

Combining ln2 with the rest of series, we have

$f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n}$
<span>
Answer: </span> $f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n}$

3 0

3 years ago

Given the domain (-2,2,4) what is the range for the relation 3x-y=3

Bumek [7]

Answer:

Given the domain, the range for 3x-y = 3 is {-9, 6, 9}

Step-by-step explanation:

First you have to put the relation in terms of y ⇒ 3x - y = 3⇒ 3x -3 = y

⇒ y = 3x - 3.

Then you replace the values indicated by the domain to find their "y" values (the ones that constitute the range).

f(-2) = -9

f(2) = 6

f(4) = 9.

Finally, the range for the given domain is {-9, 6, 9}

3 0

3 years ago

The terminal side of 0 is in quadrant II and cos 0 = -5/13. What is sin 0?

zaharov [31]

Answer:

$\displaystyle \sin\theta=\frac{12}{13}$

Step-by-step explanation:

We are given that:

$\displaystyle \cos\theta =-\frac{5}{13}\text{ where $\theta$ is in QII}$

Recall that cosine is the ratio of the adjacent side to the hypotenuse. Using the Pythagorean Theorem, solve for the opposite side (we can ignore the negative for now):

$o=\sqrt{13^2-5^2}=12$

And since θ is in QII, sine is positive, and cosine and tangent are both negative.

Sine is the ratio of the opposite side to the hypotenuse. Therefore:

$\displaystyle \sin\theta=\frac{12}{13}$

7 0

3 years ago

Read 2 more answers