2(3x-1) >=4x-6
6x-2>=4x-6
6x-4x>= -6+2
2x>=-4
x>=-2

You can see after simplification, the valve of x can be any number greater than or equal to -2. i.e -5 and -3 can not be in the solution set. the first choice can be dismissed, the fact the variable O could be any number that may not satisfy the solution. Therefore 0-1 is the answer, any thing between 0 & 1 including 0 and one satisfy the solution set

8 0

2 years ago

PLS HELP ASAP calculate the distance between the two points. (18, -24) & (87, 100)

GarryVolchara [31]

Answer:

<h3>The answer is 141.90 units</h3>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(18, -24) & (87, 100)

The distance between them is

$d = \sqrt{ ({18 - 87})^{2} + ({ - 24 - 100})^{2} } \\ = \sqrt{ ( { - 69})^{2} + ( { - 124})^{2} } \\ = \sqrt{4761 + 15376} \\ = \sqrt{20137} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = 141.90489773...$

We have the final answer as

<h3>141.90 units</h3>

Hope this helps you

5 0

3 years ago

Differentiate. <br><br> y = ln(t)/t²

azamat

Remember

the deritivive of f(x)/g(x)=(f'(x)g(x)-g'(x)f(x))/((g(x))^2)

deritivive of lnx is 1/x
derivitive of t^2=2t

so

$\frac{(1/t)t^2-2tlnt}{(t^2)^2}$ =
$\frac{t-2tlnt}{t^4}$ =
$\frac{1-2ln(t)}{t^3}$

6 0

3 years ago