All categories

3 years ago

What is the answer of (n+8)+(n-12)

Mathematics

1 answer:

notsponge [240]3 years ago

4 0

Answer:

n = 2

Step-by-step explanation:

Simplifying

(n + 8) + (n + -12) = 0

Reorder the terms:

(8 + n) + (n + -12) = 0

Remove parenthesis around (8 + n)

8 + n + (n + -12) = 0

Reorder the terms:

8 + n + (-12 + n) = 0

Remove parenthesis around (-12 + n)

8 + n + -12 + n = 0

Reorder the terms:

8 + -12 + n + n = 0

Combine like terms: 8 + -12 = -4

-4 + n + n = 0

Combine like terms: n + n = 2n

-4 + 2n = 0

Solving

-4 + 2n = 0

Solving for variable 'n'.

Move all terms containing n to the left, all other terms to the right.

Add '4' to each side of the equation.

-4 + 4 + 2n = 0 + 4

Combine like terms: -4 + 4 = 0

0 + 2n = 0 + 4

2n = 0 + 4

Combine like terms: 0 + 4 = 4

2n = 4

Divide each side by '2'.

n = 2

Simplifying

n = 2

You might be interested in

A major fishing company does its fishing in a local lake. The first year of the company's operations it managed to catch 154,000

just olya [345]

Answer:

1,116,500 Fish

Step-by-step explanation:

Pls give BRAINLIEST

8 0

2 years ago

Please Help Me

ludmilkaskok [199]

Given:

The point (9,-12) is on terminal side of angle theta in standard position.

To find:

The exact value of each of the six trigonometric functions of theta.

Solution:

The given point is (9,-12). Here, x-coordinate is positive and y-coordinate is negative. So, the point lies in 4th quadrant and only cos and sec are positive in 4th quadrant.

We know that,

$r=\sqrt{x^2+y^2}$

$r=\sqrt{9^2+(-12)^2}$

$r=\sqrt{81+144}$

$r=\sqrt{225}$

$r=15$

Now,

$\sin \theta=\dfrac{y}{r}=\dfrac{-12}{15}=-\dfrac{4}{5}$

$\cos \theta=\dfrac{x}{r}=\dfrac{9}{15}=\dfrac{3}{5}$

$\tan \theta=\dfrac{y}{x}=\dfrac{-12}{9}=-\dfrac{4}{3}$

$\cot \theta=\dfrac{1}{\tan \theta}=-\dfrac{3}{4}$

$\text{cosec} \theta=\dfrac{1}{\sin \theta}=-\dfrac{5}{4}$

$\sec \theta=\dfrac{1}{\cos \theta}=\dfrac{5}{3}$

Therefore, the values of six trigonometric functions of theta are $\sin \theta=-\dfrac{4}{5},\cos \theta=\dfrac{3}{5},\tan \theta=-\dfrac{4}{3},\cot \theta=-\dfrac{3}{4},\text{cosec} \theta=-\dfrac{5}{4},\sec \theta=\dfrac{5}{3}$ .