Please please please help! I will mark brainiest and give 20 points for answer

What are the exact values of sin(2π3radians) and cos(2π3radians)

deff fn [24]

We know that
2π/3 radians-------> convert to degrees-----> 2*180/3---> 120°
120°=90°+30°

Part a) Find <span>sin(2π/3)
</span>sin(2π/3)=sin (90°+30°)
we know that
sin (A+B)=sin A*cos B+cos A*sin B
so
sin (90°+30°)=sin 90*cos 30+cos 90*sin 30
sin 90=1
cos 30=√3/2
cos 90=0
sin 30=1/2
sin (90°+30°)=1*√3/2+0*1/2-----> √3/2

the answer part a) is
sin(2π/3)=√3/2

Part b) Find cos (2π/3)
cos (2π/3)=cos (90°+30°)
we know that
cos (A+B)=cos A*cos B-sin A*sin B
so
cos (90°+30°)=cos 90*cos 30-sin 90*sin 30
sin 90=1
cos 30=√3/2
cos 90=0
sin 30=1/2
cos (90°+30°)=0*√3/2-1*1/2----> -1/2

the answer part b) is
cos (2π/3)=-1/2

8 0

3 years ago

Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>

$x^ax^b = x^{(a+b)}$
$\dfrac{x^a}{x^b} = x^{(a-b)}$
$(a^a)^b =x^{(a\times b)}$
$(xy)^a = x^ay^a$

$x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4$

The solution of the expression is $\dfrac{4x^4}{y^6}$ .

<h2>Explanation</h2>

Given to us

$(16x^8y^{12})^{\frac{1}{2}}$

Solution

We know that 16 can be reduced to $2^4$ ,

$=(2^4x^8y^{12})^{\frac{1}{2}}$

Using identity $(xy)^a = x^ay^a$ ,

$=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}$

Using identity $(a^a)^b =x^{(a\times b)}$ ,

$=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})$

Solving further

$=2^2x^4y^{-6}$

Using identity $\dfrac{x^a}{x^b} = x^{(a-b)}$ ,

$=\dfrac{2^2x^4}{y^6}$

$=\dfrac{4x^4}{y^6}$

Hence, the solution of the expression is $\dfrac{4x^4}{y^6}$ .

Learn more about Exponentiation:

brainly.com/question/2193820

8 0

2 years ago

Given sinθ=- 3/5 and cscθ=-5/3 in quadrant III, find the value of other trigonometric functions using a Pythagorean Identity. Sh

Yakvenalex [24]

Let ABC be a triangle in the 3rd quadrant, right-angled at B.
So, AB-> Perpendicular BC -> Base AC -> Hypotenuse.
Given: sinÎ¸=-3/5 cosecÎ¸=-5/3
According to Pythagorean theorem, square of the hypotenuse is equal to the sum of square of the other two sides.
Therefore in triangle ABC, ă€–ACă€—^2=ă€–ABă€—^2+ă€–BCă€—^2 ------
--(1)
Since sinÎ¸=Perpendicular/Hypotenuse ,
AC=5 and AB=3
Substituting these values in equation (1)

ă€–BCă€—^2=ă€–ACă€—^2-ă€–ABă€—^2

ă€–BCă€—^2=5^2-3^2

ă€–BCă€—^2=25-9

ă€–BCă€—^2=16

BC=4 units

Since the triangle is in the 3rd quadrant, all trigonometric ratios, except tan
and cot are negative.
So,cosÎ¸=Base/Hypotenuse CosÎ¸=-4/5
secÎ¸=Hypotnuse/Base secÎ¸=-5/4
tanÎ¸=Perpendicular/Base tanÎ¸=3/4
cotÎ¸=Base/Perpendicular cotÎ¸=4/3

4 0

3 years ago

Helpppppppppppppppppppp plsss

insens350 [35]

Answer:

25

Step-by-step explanation:

he got an 80 so if that was his grade by answering 20 questions right then the correct answer would be 25 questions

5 0

3 years ago

What graph represents the piecewise function?

adoni [48]

Answer: Lower left corner

A piecewise function is basically a combination of other functions to make one single function. We can break up the given piecewise function into two parts:

f(x) = x-4

OR

f(x) = -2x

The f(x) will change depending on what x happens to be. If x is 0 or smaller, then we go with f(x) = x-4. Otherwise, if x is larger than 0, then we opt for f(x) = -2x.

To graph this, we basically graph y = x-4 and y = -2x together on the same coordinate system. We only graph y = x-4 if x is 0 or smaller. Likewise, we graph y = -2x when x > 0. This results in the graph shown in the lower left corner of your four answer choices.

Note: the closed circle means "include this point as part of the graph". The open circle means "exclude this point as part of the graph". So this is why the upper right corner is very close but not quite the answer we want.

5 0

3 years ago