Solve.

Mathematics

1 answer:

Bas_tet [7]2 years ago

8 0

Answer:

No Solution

Step-by-step explanation:

We are given the equation:

$\displaystyle \large{|x|-5=-20}$

First, Add both sides by 5 which leaves us $\displaystyle \large{|x|=-15}$

By definition, an absolute value cannot be negative. There are no x-values that can equal to -15 because no matter what real numbers you put in or substitute, it will never become negative.

Hence, no solution.

Read more about range at:

brainly.com/question/10197594

#SPJ1

7 0

2 years ago

Mariah025 answer here please

larisa [96]

Answer:

wht

............

..............

4 0

2 years ago

Can someone help me a still don't get this concept???

Flauer [41]

Well, if y = x - 5, then we calculate:

when x = -1 then y = -1 -5 = -6

when x = 0 then y = -5

when x = 1 then y = 1 - 5 = -4

when x = 2 then y = 2 - 5 = -3

Hope this helps!

3 0

3 years ago

Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.

Lilit [14]

For starters,

$\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k$

Consider the $n$ th partial sum, denoted by $S_n$ :

$S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n$

Multiply both sides by $\frac34$ :

$\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}$