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3 years ago

Find all solutions of the equation in the interval , 02π . =+2sinθ20 write your answer in radians in terms of π .

Mathematics

1 answer:

postnew [5]3 years ago

3 0

Hello :
2sin2θ= 0
sin2θ =0
2θ= kπ
θ= kπ/2
in [0 , 2π] :
k=0 x = π/2

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mel-nik [20]

The ans is d 2(4^2+8)

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3 years ago

What is 3·3÷3·1²·5 and for any body who doesn't get it Ur not dumb lol

r-ruslan [8.4K]

I'll give you a hint. This questions is about order of operations for pemdas. Its stand for: p-parenthesis, e-exponents, m-multiply, d-divide, a-add and s-subtract that came from left to right.

First you had to calculate with exponents first.

$1^2=1*1=1$

$3*3/3*1*5$

Then you multiply and divide from left to right.

$3*3=9$

$9/3*1*5$

$3*1*5$

$3*5$

$=15$

Final answer: $\boxed{=15}$

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

-Charlie

7 0

3 years ago

Read 2 more answers

In this exercise, we consider strings made from uppercase letters in the English alphabet and decimal digits.

Yuri [45]

Answer:

a) $26^2 36^8$

b) $21\cdot10\cdot36^7$

c) $5^3 31^7$

d) $10\cdot 9\cdot 8 \cdot 7 \cdot 26^6$

Step-by-step explanation:

We will use the product rule from combinatorics.

a) There are 26 letters in the English alphabet, so there are 26 possible choices for the first character and 26 possible choices for the last one. Each one of the remaining eight characters of the string has 36 choices (letters or digits). By the product rule, there are $26\cdot36\cdot 36\cdots 36\cdot 26=26^2 36^8$ strings.

b) We have 5 possible choices for the first character, it must be some vowel a,e,i,o,u. The second character can be chosen in 21 ways, selecting some consonant. There are 10 possibilities for the last character because only of the digits are allowed. The other seven characters have no restrictions, so each one can be chosen in 36 ways. By the product rule there are $21\cdot 10\cdot 36^7$ strings.

c) The third character has 5 possibilities. Repetition of vowels is allowed, so the sixth and eighth characters have each one 5 possible choices. There are seven characters left. None of them are a vowel, but they are allowed to take any other letter or digit, so each one of them can be chosen in 36-5=31 ways. Therefore there are $5^3 31^7$ strings.

d) Remember that the binomial coefficient $\binom{n}{k}$ is the number of ways of choosing k elements from a set of n elements. In this case, to count all the possible strings, we first need to count in how many ways we can select the four positions that will have the digits. This can be done in $\binom{10}{4}$ ways, since we are choosing four elements from the set of the ten positions of the string. Now, for the first position, we can choose any digit so it has 10 possibilities. The second position has 9 possibilities, because we can't repeat the digit used on the first position. Similarly, there are 8 choices for the third position and there are 7 choices for the fourth. Now, these are the only digits on the string, so the remaining 6 characters must be letters, then each one of them has 26 possibilities. By the product rule, there are $10\cdot 9\cdot 8 \cdot 7 \cdot 26^6$ strings.