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3 years ago

What is the value of cosine (StartFraction 4 pi Over 3 EndFraction)? cos(4pi/3)

Mathematics

2 answers:

Dominik [7]3 years ago

8 0

The answer is 1/2.....

marin [14]3 years ago

3 0

Answer:

-square root 1/2 . It is a no c. if the answer is correct plz thank

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Break the mile in to 6 pieces and run 5 parts of the mile

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3 years ago

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HELP ASAP ILL GIVE A THANKS AND BRAINLIEST

larisa [96]

Answer:

-3 < x ≤ -1

Step-by-step explanation:

we know that

The solution of the graph is equal to the interval

All real numbers greater than -3 and less than or equal to -1

so the combined inequality is equal to -3 < x ≤ -1!!!!

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3 years ago

Find the median<br> 5 12 18 7 24 16

brilliants [131]

5 7 12 16 18 24 12 and 16 are in the middle. Add 12 and 16 and then divide by 2 (28/2)= 14

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3 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago

(x+2) (y-5)=0<br> please help me graph and solve this!!

viktelen [127]

(X+2)(y-5)=0

X+2=0
X=-2

Y-5=0
Y=5

X is horizontal
Y is vertical

4 0

3 years ago