All categories

3 years ago

What is the sine value of 2 pi over 3? negative 1 over 2 1 over 2 negative square root 3 over 2 square root 3 over 2

Mathematics

1 answer:

MatroZZZ [7]3 years ago

7 0

Answer:

$\large\boxed{\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}}$

Step-by-step explanation:

$\sin\dfrac{2\pi}{3}=\sin\bigg(\pi-\dfrac{\pi}{3}\bigg)\\\\\text{use}\ \sin(x-y)=\sin x\cos y-\sin y\cos x\\\\=\sin\pi\cos\dfrac{\pi}{3}-\sin\dfrac{\pi}{3}\cos\pi\\\\\text{use the table from the attachment}\\\\\sin\pi=0\\\\\cos\dfrac{\pi}{3}=\dfrac{1}{2}\\\\\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}\\\\\cod\pi=-1\\\\\text{subtitute:}\\\\=(0)\left(\dfrac{1}{2}\right)-\left(\dfrac{\sqrt3}{2}\right)(-1)=\dfrac{\sqrt3}{2}$

You might be interested in

The sum of two numbers is 58. The larger number is 6 more than the smaller number. What are the numbers?

kenny6666 [7]

Answer:

Numbers are 32, and 26.

Step-by-step explanation:

58 / 2 = 29

29 + 3 = 32 (large number)

29 - 3 = 26 (small number)

8 0

3 years ago

Read 2 more answers

What type of solution would you get for 0=0<br> A. one solution<br> B. no solution<br> C.infinite

Otrada [13]

Answer: C. Infinite
Explanation: (To explain this answer, I'll use 1 = 1) When you get 1 = 1 after doing an equation, this mean the answer is "all real numbers", which is the same term as "infinite" because it will always be true. Therefore, 0 = 0 will always be a true statement.

8 0

3 years ago

If f(x) = 2x2 5x, find f(3b).

frez [133]

F(x) = 2x^2 + 5x
f(3b) = 2(3b)^2 + 5(3b) = 2(9b^2) + 15b = 18b^2 + 15b

7 0

3 years ago

Determine the formula for the nth term of the sequence:<br>-2,1,7,25,79,...

rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

$x_n = a n^4 + b n^3 + c n^2 + d n + e$

From the given known values of the sequence, we have

$\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}$

Solving the system yields coefficients

$a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4$

so that the n-th term in the sequence might be

$\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}$

Then the next few terms in the sequence could very well be

$\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}$

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of $\{x_n\}$ eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of $\{x_n\}$ by $\Delta^{k}\{x_n\}$ . Then

• 1st-order differences:

$\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}$

• 2nd-order differences:

$\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}$

• 3rd-order differences:

$\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}$

• 4th-order differences:

$\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}$

From here I made the assumption that $\Delta^4\{x_n\}$ is the constant sequence {15, 15, 15, …}. This implies $\Delta^3\{x_n\}$ forms an arithmetic/linear sequence, which implies $\Delta^2\{x_n\}$ forms a quadratic sequence, and so on up $\{x_n\}$ forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.

5 0

2 years ago

Please dont use Actidavid links !!!!!!!!!!!!!!!!!!!

Irina-Kira [14]

Answer: okay

Step-by-step explanation:

will do

(not use those links) ;)

5 0

3 years ago