What is the sin value of 5pi/3?

Mathematics

2 answers:

svetoff [14.1K]3 years ago

7 0

if you do not have a Unit Circle, this is a great time to get one, you can find many online just do a quick search for "unit circle".

Check the picture below.

alina1380 [7]3 years ago

6 0

Answer:

1/2

Step-by-step explanation:

Sin(5π/3) = Sin(π/3) in Q2 = 1/2

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4) Dan had 199 dollars to spend on 7 books. After

IRINA_888 [86]

Answer:

The answer is that each book costs 26 dollars

Step-by-step explanation:

First we subtract 199 dollar from 7 =182

Then we divide 182 dollars to the seven book in order to find a unit =182/7=26

To prove this answer multiply 26 dollars than 7 books and see if the outcome is 182.Then subtract 199 from 182 and see if you got 17.

3 0

3 years ago

Determine whether the equation x^3 - 3x + 8 = 0 has any real root in the interval [0, 1]. Justify your answer.

nikdorinn [45]

Answer:

The equation does not have a real root in the interval $\rm [0,1]$

Step-by-step explanation:

We can make use of the intermediate value theorem.

The theorem states that if $f$ is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:

If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
The image of a continuous function over an interval is itself an interval.

Of course, in our case, we will make use of the first one.

First, we need to proof that our function is continues in $\rm [0,1]$ , which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval $\rm [0,1]$ , which means to evaluate the equation in 0 and 1:

$f(x)=x^3-3x+8\\f(0)=8\\f(1)=6$

Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval $\rm [0,1]$ . I attached a plot of the equation in the interval $\rm [-2,2]$ where you can clearly observe how the graph does not cross the x-axis in the interval.