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babunello [35]
3 years ago
12

Find the exact values of cos (3pi/4radians) and sin (3pi/4 Radians)

Mathematics
2 answers:
-BARSIC- [3]3 years ago
8 0
Cos (3π/4) can be written as,
cos (π-π/4) = -cosπ/4 = -1/√2

similarly sin3π/4=sin(π-π/4) = sin π/4 = 1/√2

patriot [66]3 years ago
7 0

If you are doing on  apex the answer is √2/2

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A ball, thrown vertically upwards, from the ground, has its height h (in meters) expressed as a function of time t (in seconds),
Over [174]

Answer:

a. 15 meters.

b. 20 meters.

Step-by-step explanation:

a. The height of the ball at 3 seconds. 20 * 3 - 5 * (3)^2 = 60 - 5 * 9 = 60 - 45 = 15.

The ball will be 15 meters high.

b. The maximum height reached by the ball.

To get that, we need to find the vertex of the parabola. We do so by doing -b/2a to find the x-coordinate of the vertex.

In this case, a = -5 and b = 20.

-20 / 2(-5) = -20 / -10 = 20 / 10 = 2.

Then, we find the y-coordinate by putting 2 where it says "t".

h(2) =  20(2) - 5(2)^2 = (40) - 5(4) = 40 - 20 = 20 meters.

Hope this helps!

5 0
3 years ago
Read 2 more answers
Compute the sum:
Nady [450]
You could use perturbation method to calculate this sum. Let's start from:

S_n=\sum\limits_{k=0}^nk!\\\\\\\(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}

On the other hand, we have:

S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!=\\\\\\=1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!\\\\\\(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}

So from (1) and (2) we have:

\begin{cases}S_{n+1}=S_n+(n+1)!\\\\S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases}\\\\\\
S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\\\\\\
(\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}

Now, let's try to calculate sum \sum\limits_{k=0}^{n}k\cdot k!, but this time we use perturbation method.

S_n=\sum\limits_{k=0}^nk\cdot k!\\\\\\
\boxed{S_{n+1}=S_n+(n+1)(n+1)!}\\\\\\


but:

S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\=
\sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\=
\sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\
\boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}

When we join both equation there will be:

\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\
S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\
\sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\=
(n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\=
n(n+1)!+1

So the answer is:

\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}

Sorry for my bad english, but i hope it won't be a big problem :)
8 0
3 years ago
Find the area of the circle. Round your answer to the nearest hundredth.<br> 00<br> 3 in.
MAXImum [283]

Answer:

Step-by-step explanation:300

4 0
3 years ago
Write and solve an inequality that means a number<br> plus four is greater than or equal to twelve.
Stella [2.4K]
The answer is 8 because four plus eight equals twelve
4 0
3 years ago
Read 2 more answers
A standard pair of six-sided dice is rolled. What is the probability of rolling a sum less than 7? Express your answer as a frac
zzz [600]

Answer:

0.4167 is the probability of rolling a sum less than 7.  

Step-by-step explanation:

We are given the following in the question:

Event: A standard pair of six-sided dice is rolled

A: rolling a sum less than 7

Sample space:

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

A:

{(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)}

Formula:

\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}

P(A) = \dfrac{15}{36} = 0.4167

0.4167 is the probability of rolling a sum less than 7.

8 0
3 years ago
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