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3 years ago

What is the exact value of cos 5pi/4 radians?

1 answer:

3 0

Π radians = 180°
5 π / 4 = 5 · 180° / 4 = 900° / 4 = 225°
cos ( 5π/4 ) = cos 225°
With the unit circle we can find the exact value:
cos 225° = - √2 / 2 ≈ - 0.7071

You might be interested in

If 5x+6-19=5 what is x

Troyanec [42]

Answer:

<em>x </em>equals 3.6.

Step-by-step explanation:

A way to solve this equation is to preform the inverse operations on both sides.

1. Add 19 to -19 and 5: 5<em>x </em>+ 6 - 19 + 19 = 5 + 19 ---> 5x + 6 = 24

2. Subtract 6 from 6 and 5: 5x + 6 - 6 = 24 - 6 ---> 5x = 18

3. Divide 5x and 18 by 5: 5x / 5 = 18 / 5 ---> x = 3.6

4. 3.6 is the answer.

To prove it, substitute x with 3.6 in the equation: 5 · 3.6 + 6 - 19 = 5

I hope this made sense and helped you a lot! :)

5 0

3 years ago

Read 2 more answers

The area of a circle, in terms of

ch4aika [34]

Answer:

30m

Step-by-step explanation:

A=πr²

225π=πr²

225=r²

r=√225

r=15

d=2r

d=2×15

=30m

3 0

3 years ago

10 points AND BRAINLIEST EASY QUESTION MUST BE DONE IN 10 MIN OR LESS. NO TROLL ANSWERS

Ad libitum [116K]

Answer:

Initial Value=15

Rate of change=15/10=1.5

Equation: y=1.5x+15

Step-by-step explanation:

y=ax+b

y=1.5x+b

15=1.5(0)+b

b=15

3 0

2 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
\\\\\\
\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
\\\\\\
-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
\\\\\\
\textit{the initial \underline{position} was 400ft away at 0secs}
\\\\\\
-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
\\\\\\
25=t^2\implies \boxed{5=t}$

part B) check the picture below

5 0

3 years ago

PLEASE Answer! I'll Give Brainliest And A Thanks For A GOOD Answer.

melisa1 [442]

Answer: Your welcome the answer is in the attached Image.

Step-by-step explanation: mark me brainliest.

5 0

2 years ago