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Roman55 [17]
3 years ago
14

Find all solutions in the interval (0,2pi) you should get kpi and pi/4 but how do you get there

Mathematics
1 answer:
Ket [755]3 years ago
4 0

Add sin(x) to both sides

tan(x)sin(x)=sin(x)

Divide sin(x) from both sides

tan(x)=1

now you just have to figure out where tan(x)=1. you can figure this out from there by looking at the unit circle.

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Which of the following is equivalent to –2i(6 – 7i)? (0 – 2i)(6 – 7i) (0 + 2i)(6 – 7i) (–2 + i)(6 – 7i) (6 – 2i) – (7i – 2i)
Mariulka [41]

Answer:

The expression which is equivalent to –2i(6 – 7i) is; (0 – 2i)(6 – 7i)

4 0
2 years ago
Circumference of a circle of radius 13.4cm
Sedbober [7]

Answer:

26.8 pi or approx. 84.19

Step-by-step explanation:

Circumference can be found by multiplying the diameter by pi. The diameter is twice the radius, which in this case is 13.4, so it's 13.4 * 2 = 26.8.

Multiply by pi, and you've got your answer!

5 0
3 years ago
10+10 is the same as 11+11<br> do you understand?
Inga [223]

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Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
What is the answer to 9÷2/3
AlexFokin [52]

9 : 2/3 =

9 * 3/2 =

(3*9=27)

27/2 Is your answer

4 0
3 years ago
Determine whether the series is convergent or divergent. [infinity] n = 1 1 n√9 the series is a ---select--- p-series with p =
inna [77]

The series is a convergent p-series with p = 3

<h3>How to know it is a divergent or a convergent series</h3>

We would know that a series is a convergent p series when we have ∑ 1 np. Then you have to be able to tell if the series is a divergent p series or it is a convergent p series.

The way that you are able to tell this is if the terms of the series do not approach towards 0. Now when the value of p is greater than 1 then you would be able to tell that the series is a convergent series.

The value of \sqrt{9}= 3

The formular for this is

∑\frac{1}{n^p} \\

where n = 1

we know it is convergent because p is greater than 1. 3>1

Read more on convergent series here:

brainly.com/question/337693

#SPJ1

6 0
2 years ago
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