Does somebody knows how to do this problem?

Mathematics

1 answer:

pshichka [43]2 years ago

6 0

Answer:

Angle Q is 51°

Angle R is 96°

(x=5 and the exterior angle is 84°)

Step-by-step explanation:

Exterior Angle Theorem says that angle that's outside the triangle (extend a side and see the angle between the side extension and the side) IS Equal to the sum of the two angles that are farther away.

17x - 1 = 33 + 10x + 1

Get all the x's on one side of the equation.

17x-10x - 1 = 33 + 10x-10x+1

7x - 1 = 33 + 1

7x - 1 = 34 add 1 to both sides

7x = 35 divide by 7

7x/7 = 35/7

x = 5

Not finished ! Find Angle Q

10x+1 = 10(5)+1= 51°

Angle PRQ is insidethe triangle. All the angles of a triangle add up to 180°

33° + 51° + R = 180°

84° + R = 180°

R = 96°

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anyanavicka [17]

Answer:

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8$

Step-by-step explanation:

The sum you are trying to understand is this.

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}$

Remember that in general when you have a geometric series

$\sum\limits_{n = 0}^{\infty} a*r^n$ you have that

$\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}$ and that equality is true as long as $|r| < 1$ .

Therefore here we have

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n$ and $\big|\frac{-1}{15} \big| = \frac{1}{15} < 1$