All categories

3 years ago

An equation _____ has one solution. 1. never 2. always 3.sometimes

Mathematics

2 answers:

Tomtit [17]3 years ago

3 0

Answer:

Option 3

Step-by-step explanation:

An equation is sometimes one solution because it can also have infinitely many solution or no solutions.

Vilka [71]3 years ago

3 0

Answer:3

An equation can have no solutions, one solution, or a bunch of solutions

You might be interested in

Charles is planning his son’s birthday party and has a budget of $100. The Go Cart Corral charges $8 per child. How many childre

Montano1993 [528]

<h3>12 children</h3>

because

12 × 8 = 96

hope that helps :)

5 0

3 years ago

The sum of x and it’s opposite is always zero?

jeka94

Answer:

1) Let's consider the first case with the number 0 the oppose is also 0 and we have that 0-0=0 so then applies

2) Now let's consider any real number a no matter positive or negative we will have that:

$a-a =0$

Or in the other case:

$-a -(-a) =-a +a=0$

So then we can conclude that the expression is a general rule and is true

Step-by-step explanation:

For this case we can verify if the following expression is true or false:

The sum of x and it’s opposite is always zero?

If we want to proof this we need to show that for any number is true.

1) Let's consider the first case with the number 0 the oppose is also 0 and we have that 0-0=0 so then applies

2) Now let's consider any real number a no matter positive or negative we will have that:

$a-a =0$

Or in the other case:

$-a -(-a) =-a +a=0$

So then we can conclude that the expression is a general rule and is true

5 0

3 years ago

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

3 years ago

1 pump fill resovoir in 60 hrs other 80 hrs other 90 hrs when they are together how long to fill resovoir? Rsm

FromTheMoon [43]

Answer:

24.83 hours

Step-by-step explanation:

Given that :

Let time taken by pumps be :

Pump A = 60 hours

Pump B = 80 hours

Pump C = 90 hours

Recall :

Rate = 1 / time taken

The rates of the pumps are :

Pump A rate = 1 / 60

Pump B rate = 1/80

Pump C rate = 1/90

Combined rate : 1/60 + 1/80 + 1/90

Combined rate = (12 + 9 + 8) / 720 = 29/720

Combined rate = 29/720

The time taken together will be :

1 / combined rate = 1/(29/720) = 720 / 29 = 24.827 hours

8 0

3 years ago

PLEASE HURRY I DONT HAVE MUCH TIME

HACTEHA [7]

Answer:

Step-by-step explanation:

C. is correct bc it makes the most sense out of all of them just check my work by dividing by 2.

4 0

3 years ago