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Bumek [7]
3 years ago
11

Solve the math problem

Mathematics
2 answers:
Anastasy [175]3 years ago
6 0
Try 89 because a triangle adds up to 180. I think you have to do 23 +68 which is 91 then minus 91 from 180 which is 89
Lady bird [3.3K]3 years ago
4 0
23 + 68 + ? = 180
91 + ? = 180
? = 89
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Help a pimp out with this geometry
Katen [24]

Using the Pythagoras theorem since it is given that it is a right angled triangle


  let us name the triangle as A, B, C  with B as right angle

   AB^2 + BC^2 = AC^2   ( Pythagoras theorem)

   where AB = 4  , BC = 7 , AC = x

6 0
3 years ago
An equation is given. (Enter your answers as a comma-separated list. Round terms to three decimal places where appropriate. If t
pshichka [43]

Answer:

\theta = n\pi

where n = ..., -2,-1,0,1,2...

Step-by-step explanation:

given equation is:

\cos{2\theta} - 1 = 0

since no range is provided we can solve for all values of \theta:

\cos{2\theta} = 1

2\theta = \cos^{-1}{(1)}

2\theta = 0, 2\pi for one cycle of cos (0 \leq \theta \leq 2\pi)

2\theta = 0, 2\pi, 4\pi, 6\pi ... 2n\pi for all cycles of cos

we should also include negative values.

2\theta = -4\pi,-2\pi,0, 2\pi, 4\pi,... 2n\pi

we can divide each value by 2, to get the solutions for \theta instead of 2\theta

Answer:

\theta = -2\pi,-\pi,0, \pi, 2\pi,... n\pi

This is the solution of the equation \cos{2\theta} - 1 = 0.

In its most general form we can write all solutions of the equation in terms of n

\theta = n\pi where n = ..., -2,-1,0,1,2... or n is an integer.

4 0
3 years ago
Find the sum Sn below:
Hoochie [10]

We can write <em>S</em> as

\displaystyle S = \sum_{k=0}^{n-1} (n-k)3^k

and expand it as

\displaystyle S = n \sum_{k=0}^{n-1} 3^k - \sum_{k=0}^{n-1} k\cdot3^k

The first sum is geometric, nothing tricky:

\displaystyle\sum_{k=0}^{n-1} 3^k = 1 + 3 + 3^2 + \cdots + 3^{n-1} \\\\ \implies 3\sum_{k=0}^{n-1} 3^k = 3 + 3^2 + 3^3 + \cdots + 3^n \\\\ \implies -2\sum_{k=0}^{n-1} 3^k = 1 - 3^n \\\\ \implies \sum_{k=0}^{n-1} 3^k = \frac{3^n-1}2

For the second sum, you can use the same method employed in another question of yours (24494877) to find

\displaystyle \sum_{k=0}^{n-1} k\cdot 3^k = \frac{(2n-3)3^n+3}4

So this sum comes out to

\displaystyle S = n\cdot\frac{3^n-1}2 - \frac{(2n-3)3^n+3}4 \\\\ S = \frac{2n\cdot3^n-2n - (2n-3)3^n-3}4 \\\\ \boxed{S = \frac{3^{n+1}-2n-3}4}

7 0
3 years ago
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