Find the 16th term to the series PLEASEEE

Mathematics

1 answer:

slavikrds [6]2 years ago

5 0

Answer:

It is a geometric sequence.

$common \ ratio , r = \frac{12}{-4}=-3$

$a_n = a_1 \times r^{n-1} \\n = 16\\r = -3\\a_1 = -4\\a_{16 }= -4 \times (-3)^{15} = -4 \times 14348907 =57395628$

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A school is selling tickets to a drama performance. On the first day of ticket sales the school sold 3 adult tickets and 1 child

baherus [9]

Answer:

System of linear equations

$\left\{\begin{matrix} 3a+c=38 \\\\ 3a+2c=52 \end{matrix}\right.$

a: adult ticket price and c: child ticket price

Step-by-step explanation:

This system of equations can be used to find the price of the adult and child tickets.

We have two equations (one for each day) and two unknowns (adult ticket price and child ticket price).

Let a: adult ticket price and c: child ticket price,

we have for the first day that 3 adult tickets and 1 child ticket adds $38:

$3a+1c=38$

and for the second day we have that 2 adult tickets and 2 child tickets adds $52:

$3a+2c=52$

If we write this as a system of equations, we have:

$\left\{\begin{matrix} 3a+c=38 \\\\ 3a+2c=52 \end{matrix}\right.$

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3 years ago

Explain how you can tell whether the sum of two integers is positive or negative, before adding them.

julsineya [31]

Answer:

The sum of any integer and its opposite is equal to zero. adding two negative integers always yields a negative sum. To find the sum of a positive and a negative integer, take the absolute value of each integer and then subtract these values.

7 0

3 years ago

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URGENT: If θ is a second-quadrant angle and cosθ = -2/3, then tanθ = _____.

dangina [55]

In the second quadrant, both cos and tan are negative while only sin is positive.

To find tan, we will use the following property below:

$\large \boxed{ {tan}^{2} \theta = {sec}^{2} \theta - 1}$

Sec is the reciprocal of cos. If cos is a/b then sec is b/a. Since cos is 2/3 then sec is 3/2

$\large{ {tan}^{2} \theta = {( - \frac{3}{2}) }^{2} - 1} \\ \large{ {tan}^{2} \theta = \frac{9}{4} - 1} \\ \large{ {tan}^{2} \theta = \frac{9}{4} - \frac{4}{4} \longrightarrow \frac{5}{4} } \\ \large{tan \theta = \frac{ \sqrt{5} }{ \sqrt{4} } } \\ \large \boxed{tan \theta = \frac{ \sqrt{5} }{2} }$